Có $sin^2 x=\frac{1-cos 2x}{2},cos^2 2x=\frac{1+cos 4x}{2},cos^2 3x=\frac{1+cos 6x}{2}$PT ban đầu trở thành:
$1-cos 2x=2+(cos 4x+cos 6x)\Rightarrow 1+cos 2x+cos 4x+cos 6x=0$
Coi $cos 2x=cos \alpha\Rightarrow cos \alpha +cos 2\alpha +cos 3\alpha +1=0$
$\Rightarrow 4cos^3 \alpha +2cos^2\alpha -2cos \alpha =0$