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Số câu hỏi: 50
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Mã đề: #1692
Lĩnh vực: Toán học
Nhóm: Toán 10 - Cung và góc lượng giác. Công thức lượng giác
Lệ phí: Miễn phí
Lượt thi: 3318

Ôn tập trắc nghiệm Cung và góc lượng giác Toán Lớp 10 Phần 3

Câu 1

Góc có số đo \(\frac{2 \pi}{5}\) đổi sang độ là:

A.
\(240^{\circ}\)
B.
\(135^{\circ}\)
C.
\(72^{\circ}\)
D.
\(270^{\circ}\)
Câu 2

Góc có số đo \(108^0\) đổi ra rađian là:

A.
\(\frac{3 \pi}{5}\)
B.
\(\frac{\pi}{10}\)
C.
\(\frac{3 \pi}{2}\)
D.
\(\frac{\pi}{4}\)
Câu 3

Cho góc lượng giác (OA ,OB) có số đo bằng \(\frac{\pi}{5}\). Hỏi trong các số sau, số nào là số đo của một góc lượng giác có cùng tia đầu, tia cuối?

A.
\(\frac{6 \pi}{5}\)
B.
\(-\frac{11 \pi}{5}\)
C.
\(\frac{9 \pi}{5}\)
D.
\(\frac{31 \pi}{5}\)
Câu 4

Cho \(\alpha=\frac{\pi}{3}+k 2 \pi(\mathrm{k} \in \mathbb{Z}) . \text { Để } \alpha \in(19 ; 27)\) thì giá trị của k là:

A.
k=2; k=3
B.
k=4; k=3
C.
k=4; k=5
D.
k=5; k=6
Câu 5

Biết một số đo của góc \(\widehat{(O x, O y)}=\frac{3 \pi}{2}+2001 \pi\) . Giá trị tổng quát của góc \(\widehat{(Ox, Oy)}\) là:

A.
\(\widehat{(O x, O y)}=\frac{3 \pi}{2}+k \pi\)
B.
\(\widehat{(O x, O y)}=\pi+k 2 \pi\)
C.
\(\widehat{(O x, O y)}=\frac{\pi}{2}+k \pi\)
D.
\(\widehat{(O x, O y)}=\frac{\pi}{2}+k 2 \pi\)
Câu 6

Trên đường tròn lượng giác với điểm gốc là A , điểm M thuộc đường tròn sao cho cung lượng giác AM có số đo \(135^{\circ}\) . Gọi N là điểm đối xứng của M qua trục Oy , số đo cung AN là

A.
\(-45^{\circ}\)
B.
\(315^{\circ}\)
C.
\(-45^{\circ}\,\, hoặc \,\,315^{\circ}\)
D.
\(45^{\circ}+k 360^{\circ}, k \in \mathbb{Z}\)
Câu 7

Trên đường tròn lượng giác vớ điểm gốc là A . Điểm M thuộc đường tròn sao cho cung lượng giác AM có số đo \(75^0\) . Gọi N là điểm đối xứng với điểm M qua gốc tọa độ O , số đo cung lượng giác AN bằng:

A.
\(255^{0}\)
B.
\(-105^{\circ}\)
C.
\(-105^{0}\,\, hoặc\,\,255^{0}\)
D.
\(-105^{\circ}+k 360^{\circ}, k \in \mathbb{Z}\)
Câu 8

Trên đường tròn với điểm gốc là A . Điểm M thuộc đường tròn sao cho cung lượng giác AM có số đo \(60^{0}\) . Gọi N là điểm đối xứng với điểm M qua trục Oy , số đo cung AN là:

A.
\(120^{\circ}\)
B.
\(-240^{\circ}\)
C.
\(-120^{0}\,\, hoặc \,\,240^{\circ}\)
D.
\(120^{\circ}+k 360^{\circ}, k \in \mathbb{Z}\)
Câu 9

Trên đường tròn lượng giác có điểm gốc là A . Điểm M thuộc đường tròn sao cho cung lượng giác AM có số đo \(45^0\) . Gọi N là điểm đối xứng với M qua trục Ox , số đo cung lượng giác AN bằng

A.
\(-45^{0}\)
B.
\(315^{0}\)
C.
\(45^{0}\,\, hoăc \,\,315^{0}\)
D.
\(-45^{\circ}+k 360^{\circ}, k \in \mathbb{Z}\)
Câu 10

Lục giác ABCDEF nội tiếp đường tròn lượng giác có gốc là A , các đỉnh lấy theo thứ tự đó và các điểm B, C có tung độ dương. Khi đó góc lượng giác có tia đầu OA , tia cuối OC bằng:

A.
\(120^{\circ}\)
B.
\(-240^{\circ}\)
C.
\(120^{0}\,\, hoặc \,\,-240^{\circ}\)
D.
\(120^{\circ}+k 360^{\circ}, k \in \mathbb{Z}\)
Câu 11

Trên đường tròn lượng giác với điểm gốc là A , cung AN , có điểm đầu là A , điểm cuối là N 

A.
Chỉ có một số đo.
B.
Có đúng hai số đo
C.
Có đúng 4 số đo.
D.
Có vô số số đo.
Câu 12

Trên đường tròn lượng giác với điểm gốc A , cung lượng giác có số đo \(55^0\) có điểm đầu A xác định

A.
Chỉ có một điểm cuối M 
B.
Đúng hai điểm cuối M
C.
Đúng 4 điểm cuối M .
D.
Vô số điểm cuối M .
Câu 13

Trên đường tròn lượng giác, khẳng định nào sau đây đúng?

A.
Cung lượng giác có điểm đầu A và điểm cuối B chỉ có một số đo.
B.
Cung lượng giác có điểm đầu A và điểm cuối B chỉ có hai số đo sao cho tổng của chúng bằng\(2\pi\)
C.
Cung lượng giác có điểm đầu A và điểm cuối B chỉ có hai số đo hơn kém nhau \(2\pi\).
D.
Cung lượng giác có điểm đầu A và điểm cuối B có số đo sai khác nhau một bội \(2\pi\)
Câu 14

Góc lượng giác tạo bởi cung lượng giác. Trên đường tròn cung có số đo 1 rad là

A.
Cung có độ dài bằng 1
B.
Cung tương ứng với góc ở tâm \(60^0 \).
C.
Cung có độ dài bằng đường kính.
D.
Cung có độ dài bằng nửa đường kính.
Câu 15

Trên đường tròn bán kính r =15 , độ dài của cung có số đo \(50^0\) là:

A.
\(l=750\)
B.
\(l=15 \cdot \frac{180}{\pi}\)
C.
\(l=\frac{15 \pi}{180}\)
D.
\(l=15 \cdot \frac{180}{\pi} .50\)
Câu 16

Trên đường tròn bán kính r = 5 , độ dài của cung đo \(\frac{\pi}{8}\) là:

A.
\(l=\frac{\pi}{8}\)
B.
\(l=\frac{r \pi}{8}\)
C.
\(l=\frac{5 \pi}{8}\)
D.
Đáp án khác.
Câu 17

Chọn khẳng định đúng trong các khẳng định sau:

A.
\(\pi r a d=1^{0}\)
B.
\(\pi\, r a d=60^{\circ}\)
C.
\(\pi\, \mathrm{rad}=180^{\circ}\)
D.
\(\pi\, r a d=\left(\frac{180}{\pi}\right)^{0}\)
Câu 18

Chọn mệnh đề đúng

A.
\(1r a d=1^{0}\)
B.
\(1 \mathrm{rad}=60^{\circ}\)
C.
\(1 \mathrm{rad}=180^{\circ}\)
D.
\(1 \mathrm{rad}=\left(\frac{180}{\pi}\right)^{0}\)
Câu 19

Câu 1.Điểm cuối của \(\alpha \) thuộc góc phần tư thứ nhất của đường tròn lượng giác. Hãy chọn kết quả đúng

trong các kết quả sau đây

A.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaci4CaiaacM % gacaGGUbGaeqySdeMaeyOpa4JaaGimaiaac6caaaa!3CDE! \sin \alpha > 0.\)
B.
\(cos\alpha < 0\)
C.
\(tan\alpha < 0\)
D.
\(cot \alpha < 0\)
Câu 20

Nghiệm của phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaaIYaGaci4CaiaacMgacaGGUbWaaeWaaeaacaaI0aGaamiEaiab % gkHiTmaalaaabaGaeqiWdahabaGaaG4maaaaaiaawIcacaGLPaaaca % GGtaIaaGymaiabg2da9iaaicdaaaa!4394! 2\sin \left( {4x - \frac{\pi }{3}} \right)-1 = 0\) là:

A.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9maalaaabaGaeqiWdahabaGaaGioaaaacqGHRaWkcaWGRbWaaSaa % aeaacqaHapaCaeaacaaIYaaaaiaacUdacaWG4bGaeyypa0ZaaSaaae % aacaaI3aGaeqiWdahabaGaaGOmaiaaisdaaaGaey4kaSIaam4Aamaa % laaabaGaeqiWdahabaGaaGOmaaaaaaa!4A06! x = \frac{\pi }{8} + k\frac{\pi }{2};x = \frac{{7\pi }}{{24}} + k\frac{\pi }{2}\)
B.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9iaadUgacaaIYaGaeqiWdaNaai4oaiaadIhacqGH9aqpdaWcaaqa % aiabec8aWbqaaiaaikdaaaGaey4kaSIaam4AaiaaikdacqaHapaCaa % a!44F6! x = k2\pi ;x = \frac{\pi }{2} + k2\pi \)
C.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9iaadUgacqaHapaCcaGG7aGaamiEaiabg2da9iabec8aWjabgUca % RiaadUgacaaIYaGaeqiWdahaaa!436E! x = k\pi ;x = \pi + k2\pi \)
D.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9iabec8aWjabgUcaRiaadUgacaaIYaGaeqiWdaNaai4oaiaadIha % cqGH9aqpcaWGRbWaaSaaaeaacqaHapaCaeaacaaIYaaaaaaa!443A! x = \pi + k2\pi ;x = k\frac{\pi }{2}\)
Câu 21

Nghiệm của phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qaciGGJbGaai4BaiaacohacaWG4bGaey4kaSIaci4CaiaacMgacaGG % UbGaamiEaiabg2da9iaaicdaaaa!405A! \cos x + \sin x = 0\) là:

A.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9iabgkHiTmaalaaabaGaeqiWdahabaGaaGinaaaacqGHRaWkcaWG % RbGaeqiWdahaaa!3EFE! x = - \frac{\pi }{4} + k\pi \)
B.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9maalaaabaGaeqiWdahabaGaaGOnaaaacqGHRaWkcaWGRbGaeqiW % dahaaa!3E13! x = \frac{\pi }{6} + k\pi \)
C.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9iaadUgacqaHapaCaaa!3AA4! x = k\pi \)
D.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9maalaaabaGaeqiWdahabaGaaGinaaaacqGHRaWkcaWGRbGaeqiW % dahaaa!3E11! x = \frac{\pi }{4} + k\pi \)
Câu 22

Nghiệm của phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qaciGGJbGaai4BaiaacohadaahaaWcbeqaaiaaikdaaaGccaWG4bGa % ey4kaSIaci4yaiaac+gacaGGZbGaamiEaiabg2da9iaaicdaaaa!4148! {\cos ^2}x + \cos x = 0\) thỏa điều kiện: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacq % aHapaCaeaacaaIYaaaaiabgYda8iaadIhacqGH8aapdaWcaaqaaiaa % iodacqaHapaCaeaacaaIYaaaaaaa!3EC8! \frac{\pi }{2} < x < \frac{{3\pi }}{2}\)

A.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9iabec8aWbaa!39B4! x = \pi \)
B.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9maalaaabaGaeqiWdahabaGaaG4maaaaaaa!3A81! x = \frac{\pi }{3}\)
C.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9maalaaabaGaaG4maiabec8aWbqaaiaaikdaaaaaaa!3B3D! x = \frac{{3\pi }}{2}\)
D.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9iabgkHiTmaalaaabaGaaG4maiabec8aWbqaaiaaikdaaaaaaa!3C2A! x = - \frac{{3\pi }}{2}\)
Câu 23

Nghiệm của phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qaciGGJbGaai4BaiaacohadaahaaWcbeqaaiaaikdaaaGccaWG4bGa % ai4eGiGacogacaGGVbGaai4CaiaadIhacqGH9aqpcaaIWaaaaa!411D! {\cos ^2}x-\cos x = 0\) thỏa điều kiện: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGimaiabgY % da8iaadIhacqGH8aapcqaHapaCaaa!3B70! 0 < x < \pi \)

A.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9maalaaabaGaeqiWdahabaGaaGOmaaaaaaa!3A80! x = \frac{\pi }{2}\)
B.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9maalaaabaGaeqiWdahabaGaaGinaaaaaaa!3A82! x = \frac{\pi }{4}\)
C.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9maalaaabaGaeqiWdahabaGaaGOnaaaaaaa!3A84! x = \frac{\pi }{6}\)
D.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9iabgkHiTmaalaaabaGaeqiWdahabaGaaGOmaaaaaaa!3B6D! x = - \frac{\pi }{2}\)
Câu 24

Nghiệm của phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qaciGGZbGaaiyAaiaac6gapaWaaWbaaSqabeaapeGaaGOmaaaakiaa % dIhacqGHRaWkciGGZbGaaiyAaiaac6gacaWG4bGaeyypa0JaaGimaa % aa!4171! {\sin ^2}x + \sin x = 0\) thỏa điều kiện: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0YaaS % aaaeaacqaHapaCaeaacaaIYaaaaiabgYda8iaadIhacqGH8aapdaWc % aaqaaiabec8aWbqaaiaaikdaaaaaaa!3EF8! - \frac{\pi }{2} < x < \frac{\pi }{2}\)

A.
x = 0
B.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9iabec8aWbaa!39B4! x = \pi \)
C.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9maalaaabaGaeqiWdahabaGaaG4maaaaaaa!3A81! x = \frac{\pi }{3}\)
D.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9maalaaabaGaeqiWdahabaGaaGOmaaaaaaa!3A80! x= \frac{\pi }{2}\)
Câu 25

Nghiệm của phương trình \({\sin ^2}x-\sin x = 0\) thỏa điều kiện:\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGimaiabgY % da8iaadIhacqGH8aapcqaHapaCaaa!3B70! 0 < x < \pi \)

A.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9maalaaabaGaeqiWdahabaGaaGOmaaaaaaa!3A80! x = \frac{\pi }{2}\)
B.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9iabec8aWbaa!39B4! x = \pi \)
C.
x = 0
D.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9iabgkHiTmaalaaabaGaeqiWdahabaGaaGOmaaaaaaa!3B6D! x = - \frac{\pi }{2}\)
Câu 26

Nghiệm của phương trình sinx.cosx = 0 là:

A.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9maalaaabaGaeqiWdahabaGaaGOmaaaacqGHRaWkcaWGRbGaaGOm % aiabec8aWbaa!3ECB! x = \frac{\pi }{2} + k2\pi \)
B.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9iaadUgadaWcaaqaaiabec8aWbqaaiaaikdaaaaaaa!3B70! x = k\frac{\pi }{2}\)
C.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9iaadUgacaaIYaGaeqiWdahaaa!3B60! x = k2\pi \)
D.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9maalaaabaGaeqiWdahabaGaaGOnaaaacqGHRaWkcaWGRbGaaGOm % aiabec8aWbaa!3ECF! x = \frac{\pi }{6} + k2\pi \)
Câu 27

Nghiệm của phương trình sin3x = sinx là:

A.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9maalaaabaGaeqiWdahabaGaaGOmaaaacqGHRaWkcaWGRbGaeqiW % dahaaa!3E0F! x = \frac{\pi }{2} + k\pi \)
B.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9iaadUgacqaHapaCcaGG7aGaamiEaiabg2da9maalaaabaGaeqiW % dahabaGaaGinaaaacqGHRaWkcaWGRbWaaSaaaeaacqaHapaCaeaaca % aIYaaaaaaa!444C! x = k\pi ;x = \frac{\pi }{4} + k\frac{\pi }{2}\)
C.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9iaadUgacaaIYaGaeqiWdahaaa!3B60! x = k2\pi \)
D.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9maalaaabaGaeqiWdahabaGaaGOmaaaacqGHRaWkcaWGRbGaeqiW % daNaai4oaiaadUgacqGH9aqpcaWGRbGaaGOmaiabec8aWbaa!442D! x = \frac{\pi }{2} + k\pi ;k = k2\pi \)
Câu 28

Nghiệm của phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaci4yaiaac+ % gacaGGZbWaaWbaaSqabeaacaaIYaaaaOGaamiEaiabg2da9maalaaa % baGaaGymaaqaaiaaikdaaaaaaa!3D43! {\cos ^2}x = \frac{1}{2}\) là:

A.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9iabgglaXoaalaaabaGaeqiWdahabaGaaGOmaaaacqGHRaWkcaWG % RbGaaGOmaiabec8aWbaa!40B8! x = \pm \frac{\pi }{2} + k2\pi \)
B.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9maalaaabaGaeqiWdahabaGaaGinaaaacqGHRaWkcaWGRbWaaSaa % aeaacqaHapaCaeaacaaIYaaaaaaa!3EDC! x = \frac{\pi }{4} + k\frac{\pi }{2}\)
C.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9iabgglaXoaalaaabaGaeqiWdahabaGaaG4maaaacqGHRaWkcaWG % RbGaaGOmaiabec8aWbaa!40B9! x = \pm \frac{\pi }{3} + k2\pi \)
D.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9iabgglaXoaalaaabaGaeqiWdahabaGaaGinaaaacqGHRaWkcaWG % RbGaaGOmaiabec8aWbaa!40BA! x = \pm \frac{\pi }{4} + k2\pi \)
Câu 29

Phương trình 2sinx-1 = 0 có bao nhiêu nghiệm \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabgI % GiopaabmaabaGaaGimaiaacUdacaaIYaGaeqiWdahacaGLOaGaayzk % aaaaaa!3DF0! x \in \left( {0;2\pi } \right)\)?

A.
2
B.
1
C.
4
D.
vô số nghiệm
Câu 30

Số nghiệm của phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaci4yaiaac+ % gacaGGZbGaamiEaiabg2da9maalaaabaGaaGymaaqaaiaaikdaaaaa % aa!3C51! \cos x = \frac{1}{2}\) thuộc đoạn \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaamWaaeaacq % GHsislcaaIYaGaeqiWdaNaai4oaiaaikdacqaHapaCaiaawUfacaGL % Dbaaaaa!3E84! \left[ { - 2\pi ;2\pi } \right]\) là ?

A.
4
B.
2
C.
3
D.
1
Câu 31

Phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaci4CaiaacM % gacaGGUbWaaeWaaeaacaaIZaGaamiEaiabgUcaRmaalaaabaGaeqiW % dahabaGaaG4maaaaaiaawIcacaGLPaaacqGH9aqpcqGHsisldaWcaa % qaamaakaaabaGaaG4maaWcbeaaaOqaaiaaikdaaaaaaa!431C! \sin \left( {3x + \frac{\pi }{3}} \right) = - \frac{{\sqrt 3 }}{2}\) có bao nhiêu nghiệm thuộc khoảng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % aIWaGaai4oamaalaaabaGaeqiWdahabaGaaGOmaaaaaiaawIcacaGL % Paaaaaa!3B7F! \left( {0;\frac{\pi }{2}} \right)\)?

A.
3
B.
4
C.
1
D.
2
Câu 32

Số nghiệm của phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbb % a9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8frFve9Fve9 % Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaciGGZbGaaiyAai % aac6gacaaIYaGaamiEaiabg2da9maalaaabaWaaOaaaeaacaaIZaaa % leqaaaGcbaGaaGOmaaaaaaa!3CC6! \sin 2x = \frac{{\sqrt 3 }}{2}\) trong khoảng \((0;3\pi)\) là

A.
6
B.
2
C.
4
D.
1
Câu 33

Điều kiện của tham số m để phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyBaiaac6 % caciGGZbGaaiyAaiaac6gacaWG4bGaeyOeI0IaaG4maiGacogacaGG % VbGaai4CaiaadIhacqGH9aqpcaaI1aaaaa!42AC! m.\sin x - 3\cos x = 5\) có nghiệm là?

A.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaamqaaqaabe % qaaiaad2gacqGHKjYOcqGHsislcaaI0aaabaGaamyBaiabgwMiZkaa % isdaaaGaay5waaaaaa!3EBE! \left[ \begin{array}{l} m \le - 4\\ m \ge 4 \end{array} \right.\)
B.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyBaiabgw % MiZkaaisdaaaa!396A! m \ge 4\)
C.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyBaiabgw % MiZoaakaaabaGaaG4maiaaisdaaSqabaaaaa!3A42! m \ge \sqrt {34} \)
D.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0IaaG % inaiabgsMiJkaad2gacqGHKjYOcaaI0aaaaa!3CB9! - 4 \le m \le 4\)
Câu 34

Số nghiệm thuộc khoảng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbb % a9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8frFve9Fve9 % Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaadaqadaqaaiabgk % HiTiabec8aWjaacUdacqaHapaCaiaawIcacaGLPaaaaaa!3C2F! \left( { - \pi ;\pi } \right)\) của phương trình  2sinx = 1 là:

A.
1
B.
2
C.
3
D.
4
Câu 35

Tất cả các họ nghiệm của phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbbG8FasPYRqj0-yi0dXdbba9pGe9xq-JbbG8A8frFve9 % Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaaIYaGaci % 4yaiaac+gacaGGZbGaaGOmaiaadIhacqGHRaWkcaaI5aGaci4Caiaa % cMgacaGGUbGaamiEaiabgkHiTiaaiEdacqGH9aqpcaaIWaaaaa!448C! 2\cos 2x + 9\sin x - 7 = 0\) là

A.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbbG8FasPYRqj0-yi0dXdbba9pGe9xq-JbbG8A8frFve9 % Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWG4bGaey % ypa0JaeyOeI0YaaSaaaeaacqaHapaCaeaacaaIYaaaaiabgUcaRiaa % dUgacqaHapaCcaaMc8UaaGPaVpaabmaabaGaam4AaiabgIGiolabls % siIcGaayjkaiaawMcaaaaa!47EF! x = - \frac{\pi }{2} + k\pi \,\,\left( {k \in Z } \right)\)
B.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbbG8FasPYRqj0-yi0dXdbba9pGe9xq-JbbG8A8frFve9 % Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWG4bGaey % ypa0ZaaSaaaeaacqaHapaCaeaacaaIYaaaaiabgUcaRiaadUgacqaH % apaCcaaMc8UaaGPaVpaabmaabaGaam4AaiabgIGiolablssiIcGaay % jkaiaawMcaaaaa!4702! x = \frac{\pi }{2} + k\pi \,\,\left( {k \in Z } \right)\)
C.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbbG8FasPYRqj0-yi0dXdbba9pGe9xq-JbbG8A8frFve9 % Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWG4bGaey % ypa0JaeyOeI0YaaSaaaeaacqaHapaCaeaacaaIYaaaaiabgUcaRiaa % dUgacaaIYaGaeqiWdaNaaGPaVlaaykW7daqadaqaaiaadUgacqGHii % IZcqWIKeIOaiaawIcacaGLPaaaaaa!48AB! x = - \frac{\pi }{2} + k2\pi \,\,\left( {k \in Z } \right)\)
D.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbbG8FasPYRqj0-yi0dXdbba9pGe9xq-JbbG8A8frFve9 % Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWG4bGaey % ypa0ZaaSaaaeaacqaHapaCaeaacaaIYaaaaiabgUcaRiaadUgacaaI % YaGaeqiWdaNaaGPaVlaaykW7daqadaqaaiaadUgacqGHiiIZcqWIKe % IOaiaawIcacaGLPaaaaaa!47BE! x = \frac{\pi }{2} + k2\pi \,\,\left( {k \in Z } \right)\)
Câu 36

Phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiDaiaacg % gacaGGUbGaamiEaiabg2da9maakaaabaGaaG4maaWcbeaaaaa!3BA0! \tan x = \sqrt 3 \) có tập nghiệm là

A.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiWaaeaada % Wcaaqaaiabec8aWbqaaiaaiodaaaGaey4kaSIaam4AaiaaikdacqaH % apaCcaGGSaGaaGPaVlaadUgacqGHiiIZcqWIKeIOaiaawUhacaGL9b % aaaaa!4520! \left\{ {\frac{\pi }{3} + k2\pi ,\,k \in Z} \right\}\)
B.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyybIymaaa!376C! \emptyset \)
C.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiWaaeaada % Wcaaqaaiabec8aWbqaaiaaiodaaaGaey4kaSIaam4Aaiabec8aWjaa % cYcacaaMc8Uaam4AaiabgIGiolablssiIcGaay5Eaiaaw2haaaaa!4464! \left\{ {\frac{\pi }{3} + k\pi ,\,k \in Z } \right\}\)
D.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiWaaeaada % Wcaaqaaiabec8aWbqaaiaaiAdaaaGaey4kaSIaam4Aaiabec8aWjaa % cYcacaaMc8Uaam4AaiabgIGiolablssiIcGaay5Eaiaaw2haaaaa!4467! \left\{ {\frac{\pi }{6} + k\pi ,\,k \in Z } \right\}\)
Câu 37

Gọi S là tổng các nghiệm thuộc khoảng \((0;2\pi)\) của phương trình 3cosx - 1 = 0. Tính giá trị của S .

A.
S = 0
B.
\(S = 4\pi\)
C.
\(S = 3\pi\)
D.
\(S = 2\pi\)
Câu 38

Khẳng định nào sau đây là khẳng định sai?

A.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaci4yaiaac+ % gacaGGZbGaamiEaiabg2da9iabgkHiTiaaigdacqGHuhY2caWG4bGa % eyypa0JaeqiWdaNaey4kaSIaam4AaiaaikdacqaHapaCaaa!46D9! \cos x = - 1 \Leftrightarrow x = \pi + k2\pi \)
B.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaci4yaiaac+ % gacaGGZbGaamiEaiabg2da9iaaicdacqGHuhY2caWG4bGaeyypa0Za % aSaaaeaacqaHapaCaeaacaaIYaaaaiabgUcaRiaadUgacqaHapaCaa % a!45FB! \cos x = 0 \Leftrightarrow x = \frac{\pi }{2} + k\pi \)
C.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaci4yaiaac+ % gacaGGZbGaamiEaiabg2da9iaaigdacqGHuhY2caWG4bGaeyypa0Ja % am4AaiaaikdacqaHapaCaaa!434D! \cos x = 1 \Leftrightarrow x = k2\pi \)
D.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaci4yaiaac+ % gacaGGZbGaamiEaiabg2da9iaaicdacqGHuhY2caWG4bGaeyypa0Za % aSaaaeaacqaHapaCaeaacaaIYaaaaiabgUcaRiaadUgacaaIYaGaeq % iWdahaaa!46B7! \cos x = 0 \Leftrightarrow x = \frac{\pi }{2} + k2\pi \)
Câu 39

Tìm số nghiệm của phương trình sin = cos2x thuộc đoạn \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaamWaaeaaca % aIWaGaai4oaiaaikdacaaIWaGaeqiWdahacaGLBbGaayzxaaaaaa!3C92! \left[ {0;20\pi } \right]\)

A.
40
B.
30
C.
60
D.
20
Câu 40

Tìm tất cả các giá trị thực của tham số mđể phương trình sinx - m =1 có nghiệm?

A.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0IaaG % OmaiabgsMiJkaad2gacqGHKjYOcaaIWaGaaiOlaaaa!3D64! - 2 \le m \le 0.\)
B.
\(m \le 0\)
C.
\(m\ge1\)
D.
\(0\le m \le 1\)
Câu 41

tổng S của các nghiệm của phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaci4CaiaacM % gacaGGUbGaamiEaiabg2da9maalaaabaGaaGymaaqaaiaaikdaaaaa % aa!3C55! \sin x = \frac{1}{2}\)  trên đoạn \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaamWaaeaacq % GHsisldaWcaaqaaiabec8aWbqaaiaaikdaaaGaai4oamaalaaabaGa % eqiWdahabaGaaGOmaaaaaiaawUfacaGLDbaaaaa!3EA3! \left[ { - \frac{\pi }{2};\frac{\pi }{2}} \right]\)

A.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uaiabg2 % da9maalaaabaGaaGynaiabec8aWbqaaiaaiAdaaaaaaa!3B1D! S = \frac{{5\pi }}{6}\)
B.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uaiabg2 % da9maalaaabaGaeqiWdahabaGaaG4maaaaaaa!3A5B! S = \frac{\pi }{3}\)
C.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uaiabg2 % da9maalaaabaGaeqiWdahabaGaaGOmaaaaaaa!3A5A! S = \frac{\pi }{2}\)
D.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uaiabg2 % da9maalaaabaGaeqiWdahabaGaaGOnaaaaaaa!3A5E! S = \frac{\pi }{6}\)
Câu 42

Nghiệm của phương trình tan3x = tanx là

A.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9maalaaabaGaam4Aaiabec8aWbqaaiaaikdaaaGaaiilaiaaykW7 % caaMc8Uaam4AaiabgIGiolablssiIkaac6caaaa!43D3! x= \frac{{k\pi }}{2},\,\,k \in Z\)
B.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9iaadUgacqaHapaCcaGGSaGaaGPaVlaaykW7caWGRbGaeyicI4Sa % eSijHikaaa!4255! x = k\pi ,\,\,k \in Z\)
C.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9iaadUgacaaIYaGaeqiWdaNaaiilaiaaykW7caaMc8Uaam4Aaiab % gIGiolablssiIkaac6caaaa!43C3! x = k2\pi ,\,\,k \in Z\)
D.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9maalaaabaGaam4Aaiabec8aWbqaaiaaiAdaaaGaaiilaiaaykW7 % caaMc8Uaam4AaiabgIGiolablssiIkaac6caaaa!43D7! x = \frac{{k\pi }}{6},\,\,k \in Z\)
Câu 43

Nghiệm của phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaci4yaiaac+ % gacaGGZbWaaeWaaeaacaWG4bGaey4kaSYaaSaaaeaacqaHapaCaeaa % caaI0aaaaaGaayjkaiaawMcaaiabg2da9maalaaabaWaaOaaaeaaca % aIYaaaleqaaaGcbaGaaGOmaaaaaaa!416D! \cos \left( {x + \frac{\pi }{4}} \right) = \frac{{\sqrt 2 }}{2}\)là

A.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaamqaaqaabe % qaaiaadIhacqGH9aqpcaWGRbGaaGOmaiabec8aWbqaaiaadIhacqGH % 9aqpcqGHsisldaWcaaqaaiabec8aWbqaaiaaikdaaaGaey4kaSIaam % 4Aaiabec8aWbaacaGLBbaadaqadaqaaiaadUgacqGHiiIZcqWIKeIO % aiaawIcacaGLPaaaaaa!4ADE! \left[ \begin{array}{l} x = k2\pi \\ x = - \frac{\pi }{2} + k\pi \end{array} \right.\left( {k \in Z} \right)\)
B.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaamqaaqaabe % qaaiaadIhacqGH9aqpcaWGRbGaeqiWdahabaGaamiEaiabg2da9iab % gkHiTmaalaaabaGaeqiWdahabaGaaGOmaaaacqGHRaWkcaWGRbGaeq % iWdahaaiaawUfaamaabmaabaGaam4AaiabgIGiolablssiIcGaayjk % aiaawMcaaaaa!4A22! \left[ \begin{array}{l} x = k\pi \\ x = - \frac{\pi }{2} + k\pi \end{array} \right.\left( {k \in Z} \right)\)
C.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaamqaaqaabe % qaaiaadIhacqGH9aqpcaWGRbGaeqiWdahabaGaamiEaiabg2da9iab % gkHiTmaalaaabaGaeqiWdahabaGaaGOmaaaacqGHRaWkcaWGRbGaaG % Omaiabec8aWbaacaGLBbaadaqadaqaaiaadUgacqGHiiIZcqWIKeIO % aiaawIcacaGLPaaaaaa!4ADE! \left[ \begin{array}{l} x = k\pi \\ x = - \frac{\pi }{2} + k2\pi \end{array} \right.\left( {k \in Z} \right)\)
D.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaamqaaqaabe % qaaiaadIhacqGH9aqpcaWGRbGaaGOmaiabec8aWbqaaiaadIhacqGH % 9aqpcqGHsisldaWcaaqaaiabec8aWbqaaiaaikdaaaGaey4kaSIaam % 4AaiaaikdacqaHapaCaaGaay5waaWaaeWaaeaacaWGRbGaeyicI4Sa % eSijHikacaGLOaGaayzkaaaaaa!4B9A! \left[ \begin{array}{l} x = k2\pi \\ x = - \frac{\pi }{2} + k2\pi \end{array} \right.\left( {k \in Z} \right)\)
Câu 44

Nghiệm của phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaci % GGJbGaai4BaiaacohacaaMc8UaaGOmaiaadIhacqGHRaWkcaaIZaGa % ci4CaiaacMgacaGGUbGaaGPaVlaadIhacqGHsislcaaIYaaabaGaci % 4yaiaac+gacaGGZbGaaGPaVlaadIhaaaGaeyypa0JaaGimaaaa!4BDD! \frac{{\cos \,2x + 3\sin \,x - 2}}{{\cos \,x}} = 0\)là:

A.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaamqaaqaabe % qaaiaadIhacqGH9aqpdaWcaaqaaiabec8aWbqaaiaaikdaaaGaey4k % aSIaam4AaiaaikdacqaHapaCaeaacaWG4bGaeyypa0ZaaSaaaeaacq % aHapaCaeaacaaI2aaaaiabgUcaRiaadUgacqaHapaCaeaacaWG4bGa % eyypa0ZaaSaaaeaacaaI1aGaeqiWdahabaGaaGOnaaaacqGHRaWkca % WGRbGaeqiWdahaaiaawUfaaaaa!50CA! \left[ \begin{array}{l} x = \frac{\pi }{2} + k2\pi \\ x = \frac{\pi }{6} + k\pi \\ x = \frac{{5\pi }}{6} + k\pi \end{array} \right. ( k \in Z)\)
B.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaamqaaqaabe % qaaiaadIhacqGH9aqpdaWcaaqaaiabec8aWbqaaiaaiAdaaaGaey4k % aSIaam4Aaiabec8aWbqaaiaadIhacqGH9aqpdaWcaaqaaiaaiwdacq % aHapaCaeaacaaI2aaaaiabgUcaRiaadUgacqaHapaCaaGaay5waaaa % aa!47F2! \left[ \begin{array}{l} x = \frac{\pi }{6} + k\pi \\ x = \frac{{5\pi }}{6} + k\pi \end{array} \right.(k \in Z)\)
C.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaamqaaqaabe % qaaiaadIhacqGH9aqpdaWcaaqaaiabec8aWbqaaiaaikdaaaGaey4k % aSIaam4AaiaaikdacqaHapaCaeaacaWG4bGaeyypa0ZaaSaaaeaacq % aHapaCaeaacaaI2aaaaiabgUcaRiaadUgacaaIYaGaeqiWdahabaGa % amiEaiabg2da9maalaaabaGaaGynaiabec8aWbqaaiaaiAdaaaGaey % 4kaSIaam4AaiaaikdacqaHapaCaaGaay5waaaaaa!5242! \left[ \begin{array}{l} x = \frac{\pi }{2} + k2\pi \\ x = \frac{\pi }{6} + k2\pi \\ x = \frac{{5\pi }}{6} + k2\pi \end{array} \right.(k\in Z)\)
D.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaamqaaqaabe % qaaiaadIhacqGH9aqpdaWcaaqaaiabec8aWbqaaiaaiAdaaaGaey4k % aSIaam4AaiaaikdacqaHapaCaeaacaWG4bGaeyypa0ZaaSaaaeaaca % aI1aGaeqiWdahabaGaaGOnaaaacqGHRaWkcaWGRbGaaGOmaiabec8a % WbaacaGLBbaaaaa!496A! \left[ \begin{array}{l} x = \frac{\pi }{6} + k2\pi \\ x = \frac{{5\pi }}{6} + k2\pi \end{array} \right.(k\in Z)\)
Câu 45

Phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaci4CaiaacM % gacaGGUbWaaeWaaeaacaaIYaGaamiEaiabgkHiTmaalaaabaGaeqiW % dahabaGaaGinaaaaaiaawIcacaGLPaaacqGH9aqpciGGZbGaaiyAai % aac6gadaqadaqaaiaadIhacqGHRaWkdaWcaaqaaiaaiodacqaHapaC % aeaacaaI0aaaaaGaayjkaiaawMcaaaaa!4A14! \sin \left( {2x - \frac{\pi }{4}} \right) = \sin \left( {x + \frac{{3\pi }}{4}} \right)\)có tổng các nghiệm thuộc khoảng\((0;\pi)\)bằng

A.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aI3aGaeqiWdahabaGaaGOmaaaaaaa!393E! \frac{{7\pi }}{2}\)
B.
\(\pi\)
C.
\(\frac{{3\pi }}{2}\)
D.
\(\frac{{\pi }}{4}\)
Câu 46

Phương trình sin2x + 3cosx = 0 có bao nhiêu nghiệm trong khoảng \((0;\pi)\)

A.
0
B.
1
C.
2
D.
3
Câu 47

Nghiệm của phương trình 2sinx + 1 = 0  được biểu diễn trên đường tròn lượng giác ở hình bên là những điểm nào?

A.
Điểm E, điểm D .
B.
Điểm C, điểm F
C.
Điểm D, điểm C
D.
Điểm E, Điểm F
Câu 48

Cho phương trình  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Lq-Jirpepeea0-as0Fb9pgea0lXxe9vr0-vr % 0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaabaaaaaaaaape % GaaGOmaiGacohacaGGPbGaaiOBaiaadIhacqGHsisldaGcaaqaaiaa % iodaaSqabaGccqGH9aqpcaaIWaaaaa!3EA6! 2\sin x - \sqrt 3 = 0\).Tổng các nghiệm thuộc \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Lq-Jirpepeea0-as0Fb9pgea0lXxe9vr0-vr % 0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaabaaaaaaaaape % WaamWaaeaacaaIWaGaai4oaiabec8aWbGaay5waiaaw2faaaaa!3BAE! \left[ {0;\pi } \right]\)của phương trình là:

A.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaaMDbvLHfij5gC1rhimfMBNvxyNvgaCbxAamXvP5 % wqSXMqHnxAJn0BKvguHDwzZbqefqvATv2CG4uz3bIuV1wyUbqedmvE % Tj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8srps0lbb % f9q8WrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0-yr0xc9 % pIe9q8qqaq-dir-f0-yqaqVeFr0xfr-xfr-xb9adbaqaaeGaciGaai % aabeqaamaabaabaaGcbaaeaaaaaaaaa8qacqaHapaCaaa!407A! \pi\)
B.
\(\frac{\pi }{3}\)
C.
\(\frac{2\pi }{3}\)
D.
\(\frac{4\pi }{3}\)
Câu 49

Họ nghiệm của phương trình:\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaci4yaiaac+ % gacaGGZbGaamiEaiabgUcaRmaalaaabaGaaGymaaqaaiaaikdaaaGa % eyypa0JaaGimaaaa!3DED! \cos x + \frac{1}{2} = 0\)là:

A.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacq % aHapaCaeaacaaI2aaaaiabgUcaRiaadUgacaaIYaGaeqiWdahaaa!3CCC! \frac{\pi }{6} + k2\pi \)
B.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyySae7aaS % aaaeaacaaIYaGaeqiWdahabaGaaGOmaaaacqGHRaWkcaWGRbGaeqiW % dahaaa!3EB6! \pm \frac{{2\pi }}{2} + k\pi \)
C.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyySae7aaS % aaaeaacaaIYaGaeqiWdahabaGaaG4maaaacqGHRaWkcaWGRbGaaGOm % aiabec8aWbaa!3F73! \pm \frac{{2\pi }}{3} + k2\pi \)
D.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyySae7aaS % aaaeaacqaHapaCaeaacaaIZaaaaiabgUcaRiaadUgacaaIYaGaeqiW % dahaaa!3EB7! \pm \frac{\pi }{3} + k2\pi \)
Câu 50

Nghiệm của phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGOmaiGaco % hacaGGPbGaaiOBamaabmaabaGaaGinaiaadIhacqGHsisldaWcaaqa % aiabec8aWbqaaiaaiodaaaaacaGLOaGaayzkaaGaeyOeI0IaaGymai % abg2da9iaaicdaaaa!43AB! 2\sin \left( {4x - \frac{\pi }{3}} \right) - 1 = 0\)là

A.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaqcLbyacaWG4b % Gaeyypa0tcfa4aaSaaaOqaaKqzagGaeqiWdahakeaajugGbiaaiIda % aaGaey4kaSIaam4AaKqbaoaalaaakeaajugGbiabec8aWbGcbaqcLb % yacaaIYaaaaiaacUdacaaMc8UaamiEaiabg2da9Kqbaoaalaaakeaa % jugGbiaaiEdacqaHapaCaOqaaKqzagGaaGOmaiaaisdaaaGaey4kaS % Iaam4AaKqbaoaalaaakeaajugGbiabec8aWbGcbaqcLbyacaaIYaaa % aiaacYcacaaMc8Uaam4AaiabgIGiolablssiIcaa!5CA7! x = \frac{\pi }{8} + k\frac{\pi }{2};\,x = \frac{{7\pi }}{{24}} + k\frac{\pi }{2},\,k \in Z\)
B.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9maalaaabaGaeqiWdahabaGaaGioaaaacqGHRaWkcaWGRbGaaGOm % aiabec8aWjaacUdacaaMc8UaamiEaiabg2da9maalaaabaGaaG4nai % abec8aWbqaaiaaikdacaaI0aaaaiabgUcaRiaadUgacaaIYaGaeqiW % daNaaiilaiaaykW7caWGRbGaeyicI4SaeSijHikaaa!5198! x = \frac{\pi }{8} + k2\pi ;\,x = \frac{{7\pi }}{{24}} + k2\pi ,\,k \in Z\)
C.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9iaadUgacqaHapaCcaGG7aGaaGPaVlaadIhacqGH9aqpcqaHapaC % cqGHRaWkcaWGRbGaaGOmaiabec8aWjaacYcacaaMc8Uaam4AaiabgI % GiolablssiIcaa!4B20! x = k\pi ;\,x = \pi + k2\pi ,\,k \in Z\)
D.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9maalaaabaGaeqiWdahabaGaaGioaaaacqGHRaWkcaWGRbGaeqiW % daNaai4oaiaaykW7caWG4bGaeyypa0ZaaSaaaeaacaaI3aGaeqiWda % habaGaaGOmaiaaisdaaaGaey4kaSIaam4Aaiabec8aWjaacYcacaaM % c8Uaam4AaiabgIGiolablssiIcaa!5020! x = \frac{\pi }{8} + k\pi ;\,x = \frac{{7\pi }}{{24}} + k\pi ,\,k \in Z\)