Đáp án:
$\begin{array}{l}
\cos 2x - 2\sqrt 3 \sin x.\cos x = 2\sin 3x\\
\Rightarrow \cos 2x - \sqrt 3 .\sin 2x = 2\sin 3x\\
\Rightarrow \dfrac{1}{2}.\cos 2x - \dfrac{{\sqrt 3 }}{2}.\sin 2x = \sin 3x\\
\Rightarrow \sin \dfrac{\pi }{6}.\cos 2x - \cos \dfrac{\pi }{6}.sin2x = sin3x\\
\Rightarrow sin\left( {\dfrac{\pi }{6} - 2x} \right) = \sin 3x\\
\Rightarrow \left[ \begin{array}{l}
\dfrac{\pi }{6} - 2x = 3x + k2\pi \\
\dfrac{\pi }{6} - 2x = \pi - 3x + k2\pi
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{30}} - \dfrac{{k2\pi }}{5}\\
x = \dfrac{{5\pi }}{6} + k2\pi
\end{array} \right.
\end{array}$
