8.085. $4 \sin x \cos \left(\frac{\pi}{2}-x\right)+4 \sin (\pi+x) \cos x+2 \sin \left(\frac{3}{2} \pi-x\right) \cos (\pi+x)=1$.
## Solution.

By reduction formulas we have

$4 \sin x \sin x-4 \sin x \cos x+2 \cos x \cos x-1=0$

$4 \sin ^{2} x-4 \sin x \cos x+2 \cos ^{2} x-\left(\cos ^{2} x+\sin ^{2} x\right)=0$,

$3 \sin ^{2} x-4 \sin x \cos x+\cos ^{2} x=0, \Leftrightarrow 3 \operatorname{tg}^{2} x-4 \operatorname{tg} x+1=0 ;$

solving the equation as a quadratic in $\operatorname{tg} x$, we find $(\operatorname{tg} x)_{1}=\frac{1}{3}$, $x_{1}=\operatorname{arctg} \frac{1}{3}+\pi k, k \in Z ;(\operatorname{tg} x)_{2}=1, x_{2}=\frac{\pi}{4}+\pi n, n \in Z$.

Answer: $\quad x_{1}=\operatorname{arctg} \frac{1}{3}+\pi k ; x_{2}=\frac{\pi}{4}+\pi n, k, n \in Z$.