Với giá trị nào của m thì phương trình \[\sqrt 3 \sin 2x - m\cos 2x = 1\]luôn có nghiệm?

A.0     

B.1     

C.2      

D.4

A.\[m \ne \frac{1}{2}\]

B. \[m = \frac{1}{2}\]

C. \(\left\{ {\begin{array}{*{20}{c}}{\frac{1}{3} < m < 1}\\{m \ne \frac{1}{2}}\end{array}} \right.\)

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D. \[\frac{1}{3} < m < 1\]

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A.\(\left[ {\begin{array}{*{20}{c}}{x = \frac{{k\pi }}{2}}\\{x = \pm \frac{1}{2}arccos( - \frac{1}{6}) + k\pi }\end{array}} \right.(k \in Z)\)

B. \(\left[ {\begin{array}{*{20}{c}}{x = \frac{{k\pi }}{2}}\\{x = \pm \frac{5}{2}arccos( - \frac{1}{6}) + k\pi }\end{array}} \right.(k \in Z)\)

C. \(\left[ {\begin{array}{*{20}{c}}{x = \frac{{k\pi }}{2}}\\{x = \pm \frac{1}{2}arccos( - \frac{1}{3}) + k\pi }\end{array}} \right.(k \in Z)\)

D. \(\left[ {\begin{array}{*{20}{c}}{x = \frac{{k\pi }}{2}}\\{x = \pm \frac{1}{3}arccos( - \frac{1}{6}) + k\pi }\end{array}} \right.(k \in Z)\)

A.\[\left| a \right| \ge 1\]

B. \[\left| a \right| >1\]

C. \[\left| a \right| = 1\]

D. \[\left| a \right| \ne 1\]

A.0

B.1     

C.2

D.3

A.\[x = k\pi ,x = \frac{\pi }{6} + \frac{{k\pi }}{3},x = \frac{\pi }{{12}} + k\pi ,x = \frac{{5\pi }}{7} + k\pi \]

B. \[x = k2\pi ,x = \frac{{7\pi }}{6} + k2\pi ,x = - \frac{\pi }{6} + k2\pi ,x = \frac{{7\pi }}{6} + k2\pi \]

C. \[x = k\pi ,x = \frac{\pi }{3} + \frac{{k2\pi }}{3},x = \frac{\pi }{{12}} + k\pi ,x = \frac{{5\pi }}{{12}} + k\pi \]

D. \[x = k2\pi ,x = \frac{\pi }{6} + \frac{{k2\pi }}{3},x = \frac{\pi }{{12}} + k\pi ,x = \frac{{7\pi }}{{12}} + k\pi \]