Đáp án:
Giải thích các bước giải:
\[\begin{array}{l}
\frac{{(1 - 2\sin x)\cos x}}{{(1 + 2\sin x)(1 - \sin x)}} = \sqrt 3 (*)\\
Dk:\left\{ \begin{array}{l}
1 + 2\sin x \ne 0\\
1 - \sin x \ne 0
\end{array} \right. = > \left\{ \begin{array}{l}
x \ne \frac{\pi }{2} + k2\pi \\
x \ne - \frac{\pi }{6} + k2\pi \\
x \ne \frac{{7\pi }}{6} + k2\pi
\end{array} \right.\\
(*) \Leftrightarrow \frac{{\cos x - 2\sin x\cos x}}{{1 - 2{{\sin }^2}x + \sin x}} = \sqrt 3 \\
\Leftrightarrow \frac{{\cos x - \sin 2x}}{{\cos 2x + \sin x}} = \sqrt 3 \\
\Leftrightarrow \cos x - \sqrt 3 \sin x = \sin 2x + \sqrt 3 \cos 2x\\
\Leftrightarrow \cos (x + \frac{\pi }{3}) = \cos (2x - \frac{\pi }{6})\\
\Leftrightarrow \left\{ \begin{array}{l}
2x - \frac{\pi }{6} = x + \frac{\pi }{3} + k2\pi \\
2x - \frac{\pi }{6} = - x - \frac{\pi }{3} + k2\pi
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = \frac{\pi }{2} + k2\pi \\
x = - \frac{\pi }{{18}} + \frac{{k2\pi }}{3}
\end{array} \right.\\
DCDK = > {n_0}\_pt:x = - \frac{\pi }{{18}} + \frac{{k2\pi }}{3}
\end{array}\]
