11.14. Solve the equation

$$
\frac{1+\operatorname{tg} x}{1-\operatorname{tg} x}=1+\sin 2 x
$$
11.14. Answer: $x=-\frac{\pi}{4}+k \pi$ and $x=k \pi$. 

Let $t=\operatorname{tg} x$. Considering that $\sin 2 x=\frac{2 t}{1+t^{2}}$, we obtain the equation

$$
\frac{1+t}{1-t}=1+\frac{2 t}{1+t^{2}}
$$

i.e., $2(1+t) t^{2}=0$ (by the condition $t \neq 1$). This equation has roots $t_{1}=-1$ and $t_{2}=0$.