Example 3. Solve the inequality

$$
2^{y}-2 \cos x+\sqrt{y-x^{2}-1} \leqslant 0
$$
Solution. Since $y-x^{2}-1 \geqslant 0$, then $y \geqslant x^{2}+1$ and accordingly $2^{y} \geqslant 2$.

Since $\cos x \leqslant 1$, then $-2 \cos x \geqslant-2$. Adding now the inequalities of the same sense:

$$
2^{y} \geqslant 2, -2 \cos x \geqslant-2, \sqrt{y-x^{2}-1} \geqslant 0
$$

we get:

$$
2^{y}-2 \cos x+\sqrt{y-x^{2}-1} \geqslant 0
$$

Comparing this inequality with the original one, we conclude that only those values of $x$ and $y$ that satisfy $2^{y}-2 \cos x+\sqrt{y-x^{2}-1}=0$ should meet the condition of the problem, which is possible only if $2^{y}=2, \cos x=1$ and $\sqrt{y-x^{2}-1}=0$.

Solving the system:

$$
\left\{\begin{array}{l}
2^{y}=2 \\
\cos x=1 \\
\sqrt{y-x^{2}-1}=0
\end{array}\right.
$$

we get: $y=1, x=0$.