8.265. $2\left(\sin ^{6} x+\cos ^{6} x\right)-3\left(\sin ^{4} x+\cos ^{4} x\right)=\cos 2 x$.
## Solution.

We have

$2\left(\left(\sin ^{2} x\right)^{3}+\left(\cos ^{2} x\right)^{3}\right)-3\left(\left(\sin ^{2} x\right)^{2}+\left(\cos ^{2} x\right)^{2}\right)-\cos 2 x=0 \Leftrightarrow$

$\Leftrightarrow 2\left(\sin ^{2} x+\cos ^{2} x\right) \cdot\left(\sin ^{4} x-\sin ^{2} x \cos ^{2} x+\cos ^{4} x\right)-3\left(\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-\right.$

$\left.-2 \sin ^{2} x \cos ^{2} x\right)-\cos 2 x=0 \Leftrightarrow 2\left(\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-3 \sin ^{2} x \cos ^{2} x\right)-$

$-3\left(1-2 \sin ^{2} x \cos ^{2} x\right)-\cos 2 x=0 \Leftrightarrow 2-6 \sin ^{2} x \cos ^{2} x-3+6 \sin ^{2} x \times$ $\mathrm{x} \cos ^{2} x-\cos 2 x=0 \Leftrightarrow \cos 2 x=-1,2 x=\pi+2 \pi k$, $x=\frac{\pi}{2}+\pi k=\frac{\pi}{2}(2 k+1) k \in Z$.

Answer: $x=\frac{\pi}{2}(2 k+1), k \in Z$.