8.208. $\sin 6 x+2=2 \cos 4 x$.

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8.208. $\sin 6 x+2=2 \cos 4 x$.
Solution.

Write the equation as $\sin 3(2 x)+2-2 \cos 2(2 x)=0$ and applying the formulas $\sin 3 \alpha=3 \sin \alpha-4 \sin ^{3} \alpha$ and $\cos 2 \alpha=1-2 \sin ^{2} \alpha$ we get

$3 \sin 2 x-4 \sin ^{3} 2 x+2-2+4 \sin ^{2} 2 x=0 \Leftrightarrow$

$\Leftrightarrow 4 \sin ^{3} 2 x-4 \sin ^{2} 2 x-3 \sin 2 x=0 \Leftrightarrow \sin 2 x\left(4 \sin ^{2} 2 x-4 \sin 2 x-3\right)=0$.

From this

1) $\sin 2 x=0,2 x=\pi k, x_{1}=\frac{\pi k}{2}, k \in Z ;$
2) $4 \sin ^{2} 2 x-4 \sin 2 x-3=0$. Solving the equation as a quadratic in $\sin 2 x$, we have $\sin 2 x=\frac{3}{2}, \varnothing ; \sin 2 x=-\frac{1}{2}, 2 x=(-1)^{n+1} \frac{\pi}{6}+\pi n$, $x_{2}=(-1)^{n+1} \frac{\pi}{12}+\frac{\pi n}{2}, n \in Z$

Answer: $x_{1}=\frac{\pi k}{2}, x_{2}=(-1)^{n+1} \frac{\pi}{12}+\frac{\pi n}{2}$, where $k$ and $n \in Z$.