Đk: $\sin 2x \neq 0$
$\Leftrightarrow (\frac{\sin x\cos 2x-\cos x\sin 2x}{\cos x\sin 2x})\cos 4x=\frac{-1}{2}(\sin^4x+\cos^4x)$$\Leftrightarrow \frac{-\sin x}{2\cos^2x\sin x}\cos4x=\frac{-1}{2}[(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x]$
$\Leftrightarrow \cos 4x=\cos^2x(1-\frac{1}{2}\sin^22x)$
$\Leftrightarrow 2\cos^22x-1=(\frac{\cos2x+1}{2})(\frac{1}{2}+\frac{1}{2}\cos^22x)$(1)
Đặt $t = \cos 2x;\sin 2x\neq0\Rightarrow \cos 2x\neq\pm 1\Rightarrow -1<t<1$
$(1)\Leftrightarrow 2t^2-1=(\frac{t+1}{2})\frac{1}{2}(1+t^2)$
$\Leftrightarrow t^3-7t^2+t+5=0$
$\Leftrightarrow \left[ \begin{matrix} \begin{matrix} t=1(loại)\\ t=3-\sqrt{14} \end{matrix}\\ t=3+\sqrt{14}(loại)\end{matrix} \right.$