8.337. $\sin ^{4} 2 x+\sin ^{3} 2 x \cos 2 x-8 \sin 2 x \cos ^{3} 2 x-8 \cos ^{4} 2 x=0$.
## Solution.

Let's write the equation as

$\sin ^{3} 2 x(\sin 2 x+\cos 2 x)-8 \cos ^{3} 2 x(\sin 2 x+\cos 2 x)=0 \Leftrightarrow$

$\Leftrightarrow(\sin 2 x+\cos 2 x)\left(\sin ^{3} 2 x-8 \cos ^{3} 2 x\right)=0$.

From this, $\sin 2 x+\cos 2 x=0$ or $\sin ^{3} 2 x-8 \cos ^{3} 2 x=0$.

1) $\operatorname{tg} 2 x=-1,2 x=-\frac{\pi}{4}+\pi k, x_{1}=-\frac{\pi}{8}+\frac{\pi k}{2}, k \in Z$.
2) $\operatorname{tg}^{3} 2 x=8, \operatorname{tg} 2 x=2,2 x=\operatorname{arctg} 2+\pi n, x_{2}=\frac{1}{2} \operatorname{arctg} 2+\frac{\pi n}{2}, n \in Z$.

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