8.086. $\cos 6 x=2 \sin \left(\frac{3 \pi}{2}+2 x\right)$

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8.086. $\cos 6 x=2 \sin \left(\frac{3 \pi}{2}+2 x\right)$
## Solution.

Representing the equation as $\cos 3(2 x)-2 \sin \left(\frac{3 \pi}{2}+2 x\right)=0$ and applying the formula $\cos 3 \alpha=4 \cos ^{3} \alpha-3 \cos \alpha$, we have

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

## From this

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

$$
x_{2}= \pm \frac{\pi}{6}+\frac{\pi n}{2}, \quad n \in Z
$$

Answer: $\quad x_{1}=\frac{\pi}{4}(2 k+1) x_{2}= \pm \frac{\pi}{6}+\frac{\pi n}{2}, \quad k, n \in Z$