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

From the condition we have

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

## From this

1) $\operatorname{tg} 2 x+1=0, \operatorname{tg} 2 x=-1,2 x=-\frac{\pi}{4}+\pi k, x_{1}=-\frac{\pi}{8}+\frac{\pi k}{2}=\frac{\pi}{8}(4 k-1)$, $k \in Z$;
2) $\operatorname{tg}^{2} 2 x-\operatorname{tg} 2 x-2=0$. Solving this equation as a quadratic equation in $\operatorname{tg} 2 x$, we get $\operatorname{tg} 2 x=-1,2 x=-\frac{\pi}{4}+\pi l, x_{2}=-\frac{\pi}{8}+\frac{\pi l}{2}, l \in Z$;

$$
\operatorname{tg} 2 x=2,2 x=\operatorname{arctg} 2+\pi n, x_{3}=\frac{1}{2} \operatorname{arctg} 2+\frac{\pi n}{2}, n \in Z ; x_{1}=x_{2}
$$

Answer: $x_{1}=\frac{\pi}{8}(4 k-1) ; x_{2}=\frac{1}{2} \operatorname{arctg} 2+\frac{\pi n}{2}$, where $k$ and $n \in Z$.