Example 18 The solution to the equation $\left(1+x^{2}\right)^{2}=4 x\left(1-x^{2}\right)$ is $\qquad$
(2nd Zu Chongzhi Cup Junior High School Mathematics Invitational Competition)
Solution: Let $x=\operatorname{tg}^{2} \theta\left(-\frac{\pi}{2}<\theta<\frac{\pi}{2}\right)$, the original equation becomes
$$
\left(1+\operatorname{tg}^{2} \theta\right)^{2}=4 \operatorname{tg} \theta\left(1-\operatorname{tg}^{2} \theta\right) .
$$

Using $\operatorname{tg} \theta=\frac{\sin \theta}{\cos \theta}$ and $\sin ^{2} \theta+\cos ^{2} \theta=1$ to simplify, we get
$$
4 \sin \theta \cos \theta \cos 2 \theta=1
$$

or $\sin 4 \theta=1$.
$$
\begin{array}{l} 
\therefore \quad 4 \theta=2 k \pi+\frac{\pi}{2} \\
\text { or } \\
\quad \theta=\frac{k \pi}{2}+\frac{\pi}{8} . \\
\therefore \quad x=\mathrm{t}_{8!}\left(\frac{k \pi}{2}+\frac{\pi}{8}\right)(k=0,1,2,3)
\end{array}
$$