8.087. $2 \sin x \cos \left(\frac{3 \pi}{2}+x\right)-3 \sin (\pi-x) \cos x+\sin \left(\frac{\pi}{2}+x\right) \cos x=0$.
## Solution.

By reduction formulas

$$
\begin{aligned}
& 2 \sin x \sin x-3 \sin x \cos x+\cos x \cos x=0 \\
& 2 \sin ^{2} x-3 \sin x \cos x+\cos ^{2} x=0 \Leftrightarrow 2 \operatorname{tg}^{2} x-3 \operatorname{tg} x+1=0
\end{aligned}
$$

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

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