Example 1. Let's find pairs of real numbers $x$ and $y$ that satisfy the equation

$$
x^{2}+4 x \cos x y+4=0
$$
S o l u t i o n. Several methods can be proposed to solve the given equation.

1st method. Since the equation is quadratic (if we do not consider $x$ under the cosine sign), we can express $x$ in terms of trigonometric functions of the angle $x y$:

$$
x=-2 \cos x y \pm \sqrt{4 \cos ^{2} x y-4}=-2 \cos x y \pm 2 \sqrt{-\sin ^{2} x y}
$$

Since by condition $x$ and $y$ are real numbers, then $-\sin ^{2} x y \geqslant 0$. This inequality is only satisfied when $\sin x y=0$, i.e., when $x y=k \pi, k \in \mathbf{Z}$.

Now we have:

$$
x=-2 \cos x y=-2 \cos k \pi .
$$

If $k$ is even, then $\cos k \pi=1$ and $x=-2$, so $y=n \pi, n \in \mathbf{Z}$; if $k$ is odd, then $\cos k \pi=-1$ and $x=2$, so $y=\frac{\pi}{2}(2 m+$ $+1)$, where $m \in \mathbf{Z}$.

2nd method. Transform the equation so that the left side becomes a sum of non-negative numbers, for example by adding and correspondingly subtracting $4 \cos ^{2} x y$. We get:

$$
(x+2 \cos x y)^{2}+4\left(1-\cos ^{2} x y\right)=0
$$

or

$$
(x+2 \cos x y)^{2}+4 \sin ^{2} x y=0
$$

Since the sum of squares of real numbers equals zero only when each term is zero, we arrive at the system:

$$
\left\{\begin{array}{c}
x+2 \cos x y=0 \\
\sin x y=0
\end{array}\right.
$$

Solving this system, we find the same values of $x$ and $y$.