4.91 Solve the equation $\cos ^{n} x-\sin ^{n} x=1$. Here $n$ is any given natural number.
[Solution] We divide the discussion based on whether $n$ is odd or even.
(1) When $n$ is even, let $n=2 m$ (where $m$ is a positive integer). In this case, the original equation becomes
$$
\cos ^{2 m} x=1+\sin ^{2 m} x \text {. }
$$

Since $\cos ^{2 m} x \leqslant 1 \leqslant 1+\sin ^{2 m} x$,
it can only be that $\sin x=0$,
thus $\cos x= \pm 1$,
so $x=k \pi$ (where $k$ is an integer).
(2) When $n$ is odd, the original equation can be rewritten as
$$
\cos ^{n} x=1+\sin ^{n} x \text {. }
$$

Since $\sin x \geqslant-1$,
we have $\sin ^{n} x \geqslant-1$ (for $n$ being odd),
thus $\cos ^{n} x=1+\sin ^{n} x \geqslant 0$.
Also, since $n$ is odd, $\cos x \geqslant 0$.
Furthermore, since $\cos x \leqslant 1$,
we have $\cos ^{n} x \leqslant 1$,
thus $\sin ^{n} x=\cos ^{n} x-1 \leqslant 0$,
which implies $\sin x \leqslant 0$.
If $\sin x=0$, substituting into the original equation gives
$$
\cos ^{n} x=1 \text {, }
$$

which means $\cos x=1$,
thus $x=2 k \pi$ (where $k$ is an integer).
If $\sin x=-1$, substituting into the original equation gives
$$
\cos ^{n} x=0,
$$

which means $\cos x=0$,
thus $x=2 k \pi-\frac{\pi}{2}$ (where $k$ is an integer).
If $-1<\sin x<0$, in this case we have
$$
0<\cos ^{n} x<1 \text {, }
$$

which means $0<\cos x<1$,
so the original equation can be rewritten as
$$
|\cos x|^{n}+|\sin x|^{n}=1 \text {. }
$$

For $n \geqslant 3$, $|\cos x|^{n}<\cos ^{2} x,|\sin x|^{n}<\sin ^{2} x$.
Thus, $|\cos x|^{n}+|\sin x|^{n}<\sin ^{2} x+\cos ^{2} x=1$.
Therefore, for odd $n \geqslant 3$, the equation has no solution for $-1<\sin x<0$. For $n=1$, the original equation becomes
$$
\begin{array}{l}
\cos x-\sin x=1 \\
\sqrt{2} \cos \left(x+\frac{\pi}{4}\right)=1, \cos \left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2} .
\end{array}
$$

This gives $x+\frac{\pi}{4}=2 k \pi \pm \frac{\pi}{4}$,
thus $x=2 k \pi$ or $x=2 k \pi-\frac{\pi}{2}$ (where $k$ is an integer).

However, these two sets of solutions have $\sin x=0$ or $\sin x=-1$, which do not satisfy $-1<\sin x<0$. Therefore, the solutions to the original equation are
$$
x=\left\{\begin{array}{ll}
k \pi & (n \text { is even }), \\
2 k \pi \text { or } 2 k \pi-\frac{\pi}{2} & (n \text { is odd }),
\end{array}\right.
$$

where $k$ is an integer.