Điều kiện:
$\begin{array}{l}
\left\{ \begin{array}{l}
\sin x \ne - 1\\
\sin x \ne - \dfrac{1}{2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ne - \dfrac{\pi }{2} + k2\pi \\
x \ne - \dfrac{\pi }{6} + k2\pi \\
x \ne \dfrac{{7\pi }}{6} + k2\pi
\end{array} \right.
\end{array}$
$\begin{array}{l} \dfrac{{\left( {1 - 2\sin x} \right)\cos x}}{{\left( {1 + 2\sin x} \right)\left( {1 - \sin x} \right)}} = \sqrt 3 \\ \Leftrightarrow \cos x - 2\sin x\cos x = \sqrt 3 \left( {1 + 2\sin x} \right)\left( {1 - \sin x} \right)\\ \Leftrightarrow \cos x - 2\sin x\cos x = \sqrt 3 \left( {1 - \sin x + 2\sin x - 2{{\sin }^2}x} \right)\\ \Leftrightarrow \cos x - 2\sin x\cos x = \sqrt 3 \left( {1 + \sin x - 2{{\sin }^2}x} \right)\\ \Leftrightarrow \cos x - \sin 2x = \sqrt 3 \left( {\cos 2x + \sin x} \right)\\ \Leftrightarrow \cos x - \sqrt 3 \sin x = \sqrt 3 \cos 2x + \sin 2x\\ \Leftrightarrow 2\cos \left( {x + \dfrac{\pi }{3}} \right) = 2\cos \left( {2x - \dfrac{\pi }{6}} \right)\\ \Leftrightarrow \left[ \begin{array}{l} x + \dfrac{\pi }{3} = 2x - \dfrac{\pi }{6} + k2\pi \\ x + \dfrac{\pi }{3} = \dfrac{\pi }{6} - 2x + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} - x = - \dfrac{\pi }{2} + k2\pi \\ 3x = \dfrac{{ - \pi }}{6} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{\pi }{2} - k2\pi \\ x = - \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{3} \end{array} \right.\left( {k \in \mathbb{Z}} \right) \end{array}$
