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

By the formulas for reducing the degree

$\frac{1}{2}(1+\cos x)+\frac{1}{2}(1+\cos 3 x)-\frac{1}{2}(1-\cos 4 x)-\frac{1}{2}(1-\cos 8 x)=0$,

$(\cos x+\cos 3 x)+(\cos 4 x+\cos 8 x)=0 \Leftrightarrow 2 \cos 2 x \cos x+2 \cos 6 x \cos 2 x=0$, $2 \cos 2 x(\cos x+\cos 6 x)=0$.

From this:

1) $\cos 2 x=0, 2 x=\frac{\pi}{2}+\pi k, \quad x_{1}=\frac{\pi}{4}+\frac{\pi k}{2}=\frac{\pi}{4}(2 k+1), k \in Z$;
2) $\cos x+\cos 6 x=0 \Leftrightarrow 2 \cos \frac{x+6 x}{2} \cos \frac{x-6 x}{2}=0, \cos \frac{7 x}{2} \cos \frac{5 x}{2}=0 \Rightarrow$

a) $\cos \frac{7 x}{2}=0, \frac{7 x}{2}=\frac{\pi}{2}+\pi n, \quad x_{2}=\frac{\pi}{7}+\frac{2}{7} \pi n=\frac{\pi}{7}(2 n+1), n \in Z$;

b) $\cos \frac{5 x}{2}=0, \quad \frac{5 x}{2}=\frac{\pi}{2}+\pi l, \quad x_{3}=\frac{\pi}{5}+\frac{2}{5} \pi l=\frac{\pi}{5}(2 l+1), \quad l \in Z$.

Answer: $\quad x_{1}=\frac{\pi}{4}(2 k+1) x_{2}=\frac{\pi}{7}(2 n+1) x_{3}=\frac{\pi}{5}(2 l+1) \quad k, n, l \in Z$.