Đáp án:
$\begin{array}{l}
\cos 2x - \sqrt 3 \sin 2x = - 2\cos x\\
\Rightarrow - \frac{1}{2}\cos 2x + \frac{{\sqrt 3 }}{2}\sin 2x = \cos x\\
\Rightarrow - \cos \frac{\pi }{3}.\cos 2x + \sin \frac{\pi }{3}.\sin 2x = \cos x\\
\Rightarrow \cos \left( {2x + \frac{\pi }{3}} \right) = \cos x\\
\Rightarrow \left[ \begin{array}{l}
2x + \frac{\pi }{3} = x + k2\pi \\
2x + \frac{\pi }{3} = - x + k2\pi
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = - \frac{\pi }{3} + k2\pi \\
x = - \frac{\pi }{9} + \frac{{k2\pi }}{3}
\end{array} \right.
\end{array}$
