8.238. $\sin x+\cos x+\sin 2 x+\sqrt{2} \sin 5 x=\frac{2 \cot x}{1+\cot^{2} x}$.
## Solution.

Since $\sin \alpha+\cos \alpha=\sqrt{2} \cos \left(\frac{\pi}{4}-\alpha\right)$ and $\frac{2 \operatorname{ctg} \alpha}{1+\operatorname{ctg}^{2} \alpha}=\sin 2 \alpha$, we have $\sqrt{2} \cos \left(\frac{\pi}{4}-x\right)+\sqrt{2} \sin 5 x+\sin 2 x=\sin 2 x \Leftrightarrow$ $\Leftrightarrow \sqrt{2} \cos \left(\frac{\pi}{4}-x\right)+\sqrt{2} \sin 5 x=0, \cos \left(\frac{\pi}{4}-x\right)+\sin 5 x=0 \Leftrightarrow$ $\Leftrightarrow \sin \left(\frac{\pi}{4}+x\right)+\sin 5 x=0 \Leftrightarrow 2 \sin \left(3 x+\frac{\pi}{8}\right) \cos \left(2 x-\frac{\pi}{8}\right)=0$.

From this,

1) $\sin \left(3 x+\frac{\pi}{8}\right)=0,3 x+\frac{\pi}{8}=\pi k, x_{1}=-\frac{\pi}{24}+\frac{\pi k}{3}, k \in Z$;
2) $\cos \left(2 x-\frac{\pi}{8}\right)=0,2 x-\frac{\pi}{8}=\frac{\pi}{2}+\pi n, x_{2}=\frac{5 \pi}{16}+\frac{\pi n}{2}, n \in Z$.

Answer: $x_{1}=-\frac{\pi}{24}+\frac{\pi k}{3} ; x_{2}=\frac{5 \pi}{16}+\frac{\pi n}{2}$, where $k$ and $n \in Z$.