8.192. $\frac{1+\sin 2 x}{1-\sin 2 x}+2 \cdot \frac{1+\tan x}{1-\tan x}-3=0$.
## Solution.

Domain of definition: $\left\{\begin{array}{l}\sin 2 x \neq 1, \\ \operatorname{tg} x \neq 1, \\ \cos x \neq 0 .\end{array}\right.$

From the condition we have:

$$
\begin{aligned}
& \frac{\sin ^{2} x+2 \sin x \cos x+\cos ^{2} x}{\sin ^{2} x-2 \sin x \cos x+\cos ^{2} x}-2 \frac{1+\frac{\sin x}{\cos x}}{1-\frac{\sin x}{\cos x}}-3=0 \Leftrightarrow \\
& \Leftrightarrow\left(\frac{\sin x+\cos x}{\sin x-\cos x}\right)^{2}-2\left(\frac{\sin x+\cos x}{\sin x-\cos x}\right)-3=0 .
\end{aligned}
$$

Solving this equation as a quadratic equation in terms of $\frac{\sin x+\cos x}{\sin x-\cos x}$, we get

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

$\operatorname{tg} x=2, x_{2}=\operatorname{arctg} 2+\pi n, n \in Z$.

Answer: $x_{1}=\pi k ; x_{2}=\operatorname{arctg} 2+\pi n$, where $k$ and $n \in Z$.