Phương trình đã cho tương đương với:$2\sin x\cos2x=4\sin x\cos x\cos3x-\sqrt2\sin x$
$\Leftrightarrow 2\sin x\cos2x=2\sin x(\cos4x+\cos2x)-\sqrt2\sin x$
$\Leftrightarrow 2\sin x\cos4x-\sqrt2\sin x=0$
$\Leftrightarrow \sin x(2\cos4x-\sqrt2)=0$
$\Leftrightarrow \left[\begin{array}{l}\sin x=0\\\cos4x=\dfrac{\sqrt2}{2}\end{array}\right. $
$\Leftrightarrow \left[\begin{array}{l}x=k\pi\\x=\pm\dfrac{\pi}{16}+k\dfrac{\pi}{2}\end{array}\right.,k\in\mathbb{Z}$.