List $5 x_{n+1}=\frac{2 x}{1-x_{n}^{2}}, x_{1}=\sqrt{3}$. Find $x_{n}$.
Let $y_{n}=\operatorname{arc} g i_{n}$. Then $x_{n}=\operatorname{tg} y_{n}$, substituting into the original equation we get $\operatorname{tg} y_{n+1}=\operatorname{tg} 2 y_{n}$. We can take $y_{n+1}=2 y_{n}$, then $x_{n}=\operatorname{tg}\left[\frac{2^{n-1} \pi}{3}\right]$.