Giải phương trình $\sin 2x\left( {\cot x + \tan 2x} \right) = 4{\cos ^2}x$.
A. $x = \frac{\pi }{2} + k\pi ,x = \pm \frac{\pi }{6} + k\pi $, $k \in \mathbb{Z}$.
B. $x = \frac{\pi }{2} + k\pi ,x = \pm \frac{\pi }{6} + k2\pi $, $k \in \mathbb{Z}$.
C. $x = \frac{\pi }{2} + k\pi ,x = \pm \frac{\pi }{3} + k2\pi $, $k \in \mathbb{Z}$.
D. $x = \frac{\pi }{2} + k\pi ,x = \pm \frac{\pi }{3} + k\pi $, $k \in \mathbb{Z}$.
A. $x = \frac{\pi }{2} + k\pi ,x = \pm \frac{\pi }{6} + k\pi $, $k \in \mathbb{Z}$.
B. $x = \frac{\pi }{2} + k\pi ,x = \pm \frac{\pi }{6} + k2\pi $, $k \in \mathbb{Z}$.
C. $x = \frac{\pi }{2} + k\pi ,x = \pm \frac{\pi }{3} + k2\pi $, $k \in \mathbb{Z}$.
D. $x = \frac{\pi }{2} + k\pi ,x = \pm \frac{\pi }{3} + k\pi $, $k \in \mathbb{Z}$.
Chọn A.
Điều kiện: $\left\{ \begin{array}{l}\sin x \ne 0\\\cos 2x \ne 0\end{array} \right.$.
Ta có: $\sin 2x\left( {\cot x + \tan 2x} \right) = 4{\cos ^2}x$
$ \Leftrightarrow \sin 2x\left( {\frac{{\cos x}}{{\sin x.\cos 2x}}} \right) = 4{\cos ^2}x$$ \Leftrightarrow \frac{{2\sin x\cos x\cos x}}{{\sin x.\cos 2x}} = 4{\cos ^2}x$
$ \Leftrightarrow \cos x = 0 \vee \cos 2x = \frac{1}{2}$ $ \Leftrightarrow x = \frac{\pi }{2} + k\pi ,x = \pm \frac{\pi }{6} + k\pi $
Điều kiện: $\left\{ \begin{array}{l}\sin x \ne 0\\\cos 2x \ne 0\end{array} \right.$.
Ta có: $\sin 2x\left( {\cot x + \tan 2x} \right) = 4{\cos ^2}x$
$ \Leftrightarrow \sin 2x\left( {\frac{{\cos x}}{{\sin x.\cos 2x}}} \right) = 4{\cos ^2}x$$ \Leftrightarrow \frac{{2\sin x\cos x\cos x}}{{\sin x.\cos 2x}} = 4{\cos ^2}x$
$ \Leftrightarrow \cos x = 0 \vee \cos 2x = \frac{1}{2}$ $ \Leftrightarrow x = \frac{\pi }{2} + k\pi ,x = \pm \frac{\pi }{6} + k\pi $