8.240. $\operatorname{tg}\left(\frac{\pi}{4}-\frac{x}{2}\right) \cdot \frac{1+\sin x}{\sin x}=\sqrt{2} \cos x$.

8.240. $\tan\left(\frac{\pi}{4}-\frac{x}{2}\right) \cdot \frac{1+\sin x}{\sin x}=\sqrt{2} \cos x$.
## Solution.

Domain of definition: $\left\{\begin{array}{l}\sin x \neq 0, \\ \cos \left(\frac{\pi}{4}-\frac{x}{2}\right) \neq 0 .\end{array}\right.$

Since $\operatorname{tg} \frac{\alpha}{2}=\frac{1-\cos \alpha}{\sin \alpha}$, we have

$\frac{1-\cos \left(\frac{\pi}{2}-x\right)}{\sin \left(\frac{\pi}{2}-x\right)} \cdot \frac{1+\sin x}{\sin x}-\sqrt{2} \cos x=0 \Leftrightarrow$

$\Leftrightarrow \frac{1-\sin x}{\cos x} \cdot \frac{1+\sin x}{\sin x}-\sqrt{2} \cos x=0, \frac{1-\sin ^{2} x}{\cos x \sin x}-\sqrt{2} \cos x=0$,

$\frac{\cos ^{2} x}{\cos x \sin x}-\sqrt{2} \cos x=0, \cos ^{2} x-\sqrt{2} \cos ^{2} x \sin x=0$,

$\cos ^{2} x(1-\sqrt{2} \sin x)=0$.

From this,

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

Answer: $x_{1}=\frac{\pi}{2}(4 k+1) ; x_{2}=(-1)^{n} \frac{\pi}{4}+\pi n$, where $k$ and $n \in Z$.