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

By the formulas for reducing the power

$$
\begin{aligned}
& \frac{1}{2}(1+\cos 2 x)+\frac{1}{2}(1+\cos 4 x)-\frac{1}{2}(1+\cos 6 x)-\frac{1}{2}(1+\cos 8 x)=0 \\
& (\cos 2 x+\cos 4 x)-(\cos 6 x+\cos 8 x)=0 \Leftrightarrow 2 \cos 3 x \cos x-2 \cos 7 x \cos x=0 \\
& 2 \cos x(\cos 3 x-\cos 7 x)=0
\end{aligned}
$$

From this, either $\cos x=0$, or $\cos 3 x-\cos 7 x=0$. From the first equation, we have $x_{1}=\frac{\pi}{2}+\pi k, k \in \mathbb{Z}$. The second equation is equivalent to the following: $\sin 5 x \sin 2 x=0$, from which either $\sin 5 x=0, 5 x=\pi n, x_{2}=\frac{\pi n}{5}, n \in \mathbb{Z}$, or $\sin 2 x=0, 2 x=\pi l, x_{3}=\frac{\pi l}{2}, l \in \mathbb{Z}$; $x_{1}$ is included in $x_{3}$.

Answer: $\quad x_{1}=\frac{\pi n}{5} ; x_{2}=\frac{\pi l}{2}, n, l \in \mathbb{Z}$.