8.032. $0.5(\cos 5 x+\cos 7 x)-\cos ^{2} 2 x+\sin ^{2} 3 x=0$.
## Solution.

## Rewrite the equation as

$\cos \frac{5 x+7 x}{2} \cos \frac{5 x-7 x}{2}-\frac{1}{2}(1+\cos 4 x)+\frac{1}{2}(1-\cos 6 x)=0 \Leftrightarrow$

$\Leftrightarrow 2 \cos 6 x \cos x-1-\cos 4 x+1-\cos 6 x=0, 2 \cos 6 x \cos x-$

$-(\cos 4 x+\cos 6 x)=0 \Leftrightarrow 2 \cos 6 x \cos x-2 \cos \frac{4 x+6 x}{2} \cos \frac{4 x-6 x}{2}=0$,

$\cos 6 x \cos x-\cos 5 x \cos x=0 \Leftrightarrow \cos x(\cos 6 x-\cos 5 x)=0$.

From this:

1) $\cos x=0, \quad x_{1}=\frac{\pi}{2}+\pi k=\frac{\pi}{2}(2 k+1), k \in Z$;
2) $\cos 6 x-\cos 5 x=0,-2 \sin \frac{6 x+5 x}{2} \sin \frac{5 x-6 x}{2}=0$.

Then $\sin \frac{11}{2} x=0, \quad \frac{11}{2} x=\pi n, x_{2}=\frac{2 \pi n}{11}, n \in Z ; \sin \frac{x}{2}=0, \frac{x}{2}=\pi l$, $x_{3}=2 \pi l, \quad l \in Z ; x_{3}$ is included in $x_{2}$.

Answer: $x_{1}=\frac{\pi}{2}(2 k+1) x_{2}=\frac{2 \pi n}{11}, k, n \in Z$.