8.109. $\cos 2 x+\cos 6 x+2 \sin ^{2} x=1$.
## Solution.

From the condition

$\cos 2 x+\cos 6 x-\left(1-2 \sin ^{2} x\right)=0 \Leftrightarrow \cos 2 x+\cos 6 x-\cos 2 x=0, \cos 6 x=0$.

Then $6 x=\frac{\pi}{2}+\pi n, x=\frac{\pi}{12}+\frac{\pi n}{6}=\frac{\pi}{12}(2 n+1), n \in Z$.

Answer: $\quad x=\frac{\pi}{12}(2 n+1), \quad n \in Z$.