8.278. $4 \sin ^{3} x \cos 3 x+4 \cos ^{3} x \sin 3 x=3 \sin 2 x$.

8.278. $4 \sin ^{3} x \cos 3 x + 4 \cos ^{3} x \sin 3 x = 3 \sin 2 x$.
## Solution.

Let's rewrite the equation as

$2 \sin ^{2} x(2 \sin x \cos 3 x)+2 \cos ^{2} x(2 \sin 3 x \cos x)-3 \sin 2 x=0 \Leftrightarrow$

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

$\Leftrightarrow\left(2 \cos ^{2} x \sin 2 x-2 \sin ^{2} x \sin 2 x\right)+\left(2 \sin ^{2} x \sin 4 x+2 \cos ^{2} x \sin 4 x\right)-$ $-3 \sin 2 x=0 \Leftrightarrow 2 \sin 2 x\left(\cos ^{2} x-\sin ^{2} x\right)+2 \sin 4 x\left(\sin ^{2} x+\cos ^{2} x\right)-$ $-3 \sin 2 x=0 \Leftrightarrow 2 \sin 2 x \cos 2 x+2 \sin 4 x-3 \sin 2 x=0 \Leftrightarrow$ $\Leftrightarrow \sin 4 x-\sin 2 x=0$. From this,

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

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