Đáp án:
$\begin{array}{l}
{\left( {\cos 2x - sin2x} \right)^2} + 2\left( {\sin 3x - \sin x} \right).\cos x - 1 = 0\\
\Rightarrow {\cos ^2}2x - 2\cos 2x.sin2x + si{n^2}2x + 2.2.\cos 2x.\sin x.\cos x - 1 = 0\\
\Rightarrow {\cos ^2}2x - 2\cos 2x.sin2x + si{n^2}2x - 2.cos2x.sin2x - 1 = 0\\
\Rightarrow {\cos ^2}2x + {\sin ^2}2x - 1 = 0\\
\Rightarrow 1 - 1 = 0\\
\Rightarrow 0 = 0\left( {luôn\,đúng} \right)
\end{array}$
