8.067. $\sin 3 x+\sin 5 x=2\left(\cos ^{2} 2 x-\sin ^{2} 3 x\right)$.
Solution.

Rewrite the given equation as

$2 \sin \frac{3 x+5 x}{2} \cos \frac{3 x-5 x}{2}=2\left(\frac{1}{2}(1+\cos 4 x)-\frac{1}{2}(1-\cos 6 x)\right) \Leftrightarrow$

$2 \sin 4 x \cos x=\cos 4 x+\cos 6 x \Leftrightarrow$

$\Leftrightarrow 2 \sin 4 x \cos x-2 \cos \frac{4 x+6 x}{2} \cos \frac{4 x-6 x}{2}=0 \Leftrightarrow$

$\Leftrightarrow \sin 4 x \cos x-\cos 5 x \cos x=0, \cos x(\sin 4 x-\cos 5 x)=0$.

## From this

1) $\cos x=0, \quad x_{1}=\frac{\pi}{2}+\pi n=\frac{\pi}{2}(2 n+1), \quad n \in Z$
2) $\sin 4 x-\cos 5 x=0 \Leftrightarrow 2 \sin \frac{4 x-\frac{\pi}{2}+5 x}{2} \cos \frac{4 x+\frac{\pi}{2}-5 x}{2}=0$, $\sin \left(\frac{9 x}{2}-\frac{\pi}{4}\right) \cos \left(\frac{x}{2}-\frac{\pi}{4}\right)=0$

Then either $\sin \left(\frac{9 x}{2}-\frac{\pi}{4}\right)=0, \frac{9 x}{2}-\frac{\pi}{4}=\pi k, \frac{9 x}{2}=\frac{\pi}{4}+\pi k, \quad x_{2}=\frac{\pi}{18}+$ $+\frac{2}{9} \pi k=\frac{\pi}{18}(4 k+1), k \in Z$, or $\cos \left(\frac{x}{2}-\frac{\pi}{4}\right)=0, \frac{x}{2}-\frac{\pi}{4}=\frac{\pi}{2}+\pi l, \frac{x}{2}=\frac{3 \pi}{4}+$ $+\pi, x_{3}=\frac{3}{2} \pi+2 \pi l, l \in Z ; x_{3}$ is included in $x_{1}$.

Answer: $\quad x_{1}=\frac{\pi}{2}(2 n+1) x_{2}=\frac{\pi}{18}(4 k+1), n, k \in Z$.