【Example 3.42】If $P_{1}(x)=x^{2}-2, P_{i}(x)=P_{1}\left[P_{i-1}(x)\right], i=$ $2,3, \cdots, n$. Solve $P_{n}(x)=x$.
To solve: By induction, it is easy to prove that if $|x|>2$, then $P_{n}(x)>x$, i.e., there is no solution. Therefore, we only need to discuss $|x| \leqslant 2$, and we can let $x=2 \cos t$.
Note: $2 \cos ^{2} t-1=\cos 2 t$, by induction it is easy to prove
$$
\begin{array}{c}
P_{n}(2 \cos t)=2 \cos 2^{n} t . \\
\therefore P_{n}(x)=x,
\end{array}
$$

i.e.,
$$
\begin{array}{c}
\cos 2^{n} t=\cos t, \\
\therefore \sin \frac{2^{n}-1}{2} t \sin \frac{2^{n}+1}{2} t=0 .
\end{array}
$$

Thus, $P_{n}(x)=x$ can be transformed into a $2^{n}$-degree equation, with $2^{n}$ solutions:
$$
x=2 \cos \frac{2 m \pi}{2^{n}-1}, \quad m=0,1,3, \cdots, 2^{n-1}-1,
$$

and
$$
x=2 \cos \frac{2 k \pi}{2^{n}+1}, \quad k=0,1,3, \cdots, 2^{n-1}-1 .
$$