Ta có:
$2\cos x\cos2x\cos3x-7=7\cos2x$
$\Leftrightarrow \cos2x(\cos4x+\cos2x)-7-7\cos2x=0$
$\Leftrightarrow \cos2x(2\cos^22x-1+\cos2x)-7-7\cos2x=0$
$\Leftrightarrow 2\cos^32x+\cos^22x-8\cos2x-7=0$
$\Leftrightarrow (\cos2x+1)(2\cos^22x-\cos2x-7)=0$
$\Leftrightarrow \left[\begin{array}{l}\cos2x=-1\\\cos2x=\dfrac{1\pm\sqrt{57}}{4}\end{array}\right.$
$\Leftrightarrow \cos2x=-1 \Leftrightarrow x=\dfrac{\pi}{2}+k\pi.k\in\mathbb{Z}$