$\Leftrightarrow \frac{\sin 2x\cos x-\cos 2x\sin x}{\cos x\cos 2x}=\frac{1}{3}\cos x(3\sin x-4sin^{3}x)$
$\Leftrightarrow \frac{\sin x}{(2cos^{2}x-1)\cos x}=\frac{1}{3}cosx\sin x(3-4(1-cos^{2}x))$
$\Leftrightarrow 3=cos^{2}x.(2cos^{2}x-1).(4cos^{2}x-1) (1) hoặc \sin x=0$
$\sin x=0\rightarrow x=k\pi $
$(1)\Leftrightarrow 8cos^{6}x-6cos^{4}x+cos^{2}x-3=0$
$\Leftrightarrow (cos^{2}x-1)(8cos^{4}x+2cos^{2}x+3)=0\Leftrightarrow \cos x=\pm 1\rightarrow x=k\pi $