8.212. $\cos 9 x-2 \cos 6 x=2$.

8.212. $\cos 9 x-2 \cos 6 x=2$. 

(Note: The equation is already in a universal mathematical format and does not change in translation.)
## Solution.

Let's write the equation as $\cos 3(3 x)-2 \cos 2(3 x)-2=0$ and, applying the formulas $\cos 3 \alpha=4 \cos ^{3} \alpha-3 \cos \alpha$ and $\cos 2 \alpha=2 \cos ^{2} \alpha-1$, we have:

$$
\begin{aligned}
& 4 \cos ^{3} 3 x-3 \cos 3 x-2\left(2 \cos ^{2} 3 x-1\right)-2=0 \Leftrightarrow \\
& \Leftrightarrow 4 \cos ^{3} 3 x-4 \cos ^{2} 3 x-3 \cos 3 x=0 \Leftrightarrow \cos 3 x\left(4 \cos ^{2} 3 x-4 \cos 3 x-3\right)=0
\end{aligned}
$$

From this,

1) $\cos 3 x=0, 3 x=\frac{\pi}{2}+\pi k, x_{1}=\frac{\pi}{6}+\frac{\pi k}{3}=\frac{\pi}{6}(2 k+1), k \in Z$,
2) $4 \cos ^{2} 3 x-4 \cos 3 x-3=0$. Solving this equation as a quadratic in $\cos 3 x$, we find $\cos 3 x=\frac{3}{2}, \varnothing$; or $\cos 3 x=-\frac{1}{2}, 3 x= \pm \frac{2}{3} \pi+2 \pi n$,

$$
x_{2}= \pm \frac{2}{9} \pi+\frac{2}{3} \pi n=\frac{2 \pi}{9}(3 n \pm 1), n \in Z
$$

Answer: $x_{1}=\frac{\pi}{6}(2 k+1), x_{2}=\frac{2 \pi}{9}(3 n \pm 1)$, where $k$ and $n \in Z$.