8.344. $\cos 2 x=\cos ^{2} \frac{3 x}{2}$.

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8.344. $\cos 2 x=\cos ^{2} \frac{3 x}{2}$.
## Solution.

From the condition we have

$$
\cos 2 x=\frac{1}{2}(1+\cos 3 x)
$$

and using the formulas $\cos 2 \alpha=2 \cos ^{2} \alpha-1, \cos 3 x=4 \cos ^{3} x-3 \cos x$, we get

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

## From this

1) $\cos x-1=0, \cos x=1, x_{1}=2 \pi k, k \in Z$;
2) $4 \cos ^{3} x-3=0, \cos x= \pm \frac{\sqrt{3}}{2}, x_{2}= \pm \frac{\pi}{6}+\pi n=\frac{\pi}{6}(6 n \pm 1), n \in Z$.

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