$2cos^{2}2x+cos2x=4sin^{2}2x.cos^{2}x$
$\Leftrightarrow 2cos^{2}2x+cos2x=4sin^{2}2x .\dfrac{1+\cos 2x}{2}$
$\Leftrightarrow 4\cos^2 2x +2\cos 2x = 4(1-\cos^2 2x)(1+\cos 2x)$ đặt $\cos 2x = t;\ t \in [-1;\ 1]$
$\Leftrightarrow 4t^2 +2t = 4(1-t^2)(1+t)$
$\Leftrightarrow 2t^3 +4t^2 -t -2=0$
+ $t = -2$ loại
+ $t =-\dfrac{1}{\sqrt 2} =\cos 2x \Rightarrow 2x = \pm \dfrac{3\pi}{4} +k2\pi$
+ $t =\dfrac{1}{\sqrt 2} =\cos 2x \Rightarrow 2x = \pm \dfrac{\pi}{4} +k2\pi$
Coi như xong nhé, chia 2 sang là ok