Đáp án:
$\begin{array}{l}
{\left( {\sin 2x + \sqrt 3 \cos 2x} \right)^2} = 2\cos \left( {2x - \dfrac{\pi }{6}} \right)\\
\Rightarrow {\left( {\dfrac{1}{2}\sin 2x + \dfrac{{\sqrt 3 }}{2}\cos 2x} \right)^2} = \dfrac{1}{2}\cos \left( {2x - \dfrac{\pi }{6}} \right)\\
\Rightarrow {\left( {\sin \dfrac{\pi }{6}.\sin 2x + \cos \dfrac{\pi }{6}.\cos 2x} \right)^2} = \dfrac{1}{2}.\cos \left( {2x - \dfrac{\pi }{6}} \right)\\
\Rightarrow {\cos ^2}\left( {2x - \dfrac{\pi }{6}} \right) = \dfrac{1}{2}\cos \left( {2x - \dfrac{\pi }{6}} \right)\\
\Rightarrow \left[ \begin{array}{l}
\cos \left( {2x - \dfrac{\pi }{6}} \right) = 0\\
\cos \left( {2x - \dfrac{\pi }{6}} \right) = \dfrac{1}{2}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
2x - \dfrac{\pi }{6} = \dfrac{\pi }{2} + k\pi \\
2x - \dfrac{\pi }{6} = \dfrac{\pi }{3} + k2\pi \\
2x - \dfrac{\pi }{6} = - \dfrac{\pi }{3} + k2\pi
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{3} + \dfrac{{k\pi }}{2}\\
x = \dfrac{\pi }{4} + k\pi \\
x = - \dfrac{\pi }{{12}} + k\pi
\end{array} \right.
\end{array}$
