8.022. $2 \sin ^{3} x+2 \sin ^{2} x \cos x-\sin x \cos ^{2} x-\cos ^{3} x=0$.
Solution.

Rewrite the equation as

$$
\begin{aligned}
& 2 \sin ^{2} x(\sin x+\cos x)-\cos ^{2} x(\sin x+\cos x)=0, \Leftrightarrow \\
& \Leftrightarrow(\sin x+\cos x)\left(2 \sin ^{2} x-\cos ^{2} x\right)=0 .
\end{aligned}
$$

Then:

1) $\sin x+\cos x=0$

or
2) $2 \sin ^{2} x-\cos ^{2} x=0$.

We have:

1) $\operatorname{tg} x=-1$

or
2) $\operatorname{tg}^{2} x=\frac{1}{2}$, from which $x_{1}=-\frac{\pi}{4}+\pi n=\frac{\pi}{4}(4 n-1), n \in Z$; $x_{2,3}= \pm \operatorname{arctg} \frac{\sqrt{2}}{2}+\pi k, k \in Z$.

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