8.154. $\sin 6 x+\sin 2 x=\frac{1}{2} \operatorname{tg} 2 x$.

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8.154. $\sin 6 x+\sin 2 x=\frac{1}{2} \tan 2 x$.
## Solution.

Domain of definition: $\cos 2 x \neq 0$.

Using the formula $\sin \alpha+\sin \beta=2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}$, we get

$2 \sin \frac{6 x+2 x}{2} \cos \frac{6 x-2 x}{2}-\frac{\sin 2 x}{2 \cos 2 x}=0, 2 \sin 4 x \cos 2 x-\frac{\sin 2 x}{2 \cos 2 x}=0$, $4 \sin 2 x \cos 2 x \cos 2 x-\frac{\sin 2 x}{2 \cos 2 x}=0 \Rightarrow 8 \sin 2 x \cos ^{3} 2 x-\sin 2 x=0$, $\sin 2 x\left(8 \cos ^{3} 2 x-1\right)=0$.

From this:

1) $\sin 2 x=0, 2 x=\pi n, \quad x_{1}=\frac{\pi n}{2}, n \in Z$
2) $8 \cos ^{3} 2 x-1=0, \cos 2 x=\frac{1}{2}, 2 x= \pm \frac{\pi}{3}+2 \pi k, x_{2}= \pm \frac{\pi}{6}+\pi k=\frac{\pi}{6}(6 k \pm 1)$, $k \in Z$

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