$2\cos2x=\sin3x-\sin x$
$\to 2\cos2x=2\cos2x.\sin x$
$\to 2\cos2x(\sin x-1)=0$
$\to \left[ \begin{array}{l}\cos2x=0\\\sin x=1\end{array} \right.$ $\to \left[ \begin{array}{l}x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x=\dfrac{\pi}{2}+k2\pi\end{array} \right.$