【Example 4.1】Find the real solutions of the following equation:
$$
x^{2}+2 x \sin (x y)+1=0 \text {. }
$$
$$
\begin{array}{l}
\because[x+\sin (x y)]^{2}+\cos ^{2}(x y)=0 \\
\therefore \cos (x y)=0, \quad x=-\sin (x y)
\end{array}
$$

From $\cos (x y)=0$, we get
$$
x y=n \pi+\frac{\pi}{2} .
$$

Thus, the solution is
$$
\begin{array}{l}
x=-\sin (x y)=-\sin \left(n \pi+\frac{\pi}{2}\right)=(-1)^{n+1} \\
y=(-1)^{n+1}\left(n \pi+\frac{\pi}{2}\right) . \quad(n=0, \pm 1, \cdots)
\end{array}
$$