2. Solve the equation: $\cos ^{4}\left(\sqrt{\frac{\pi^{2}}{4}-|x|}\right)+\sin ^{4}\left(\sqrt{\frac{\pi^{2}}{4}-|x|}\right)=\sin ^{-2}\left(\sqrt{\frac{\pi^{2}}{4}-|x|}\right)$.
# Solution.

Let's make the substitution $t=\left(\sqrt{\frac{\pi^{2}}{4}-|x|}\right)$. Clearly, $0<t \leq \frac{\pi}{2}$.

$\cos ^{4} t+\sin ^{4} t=\sin ^{-2} t$.

$\cos ^{4} t+\sin ^{4} t=\left(\cos ^{2} t+\sin ^{2} t\right)^{2}-2 \sin ^{2} t \cdot \cos ^{2} t=1-2 \sin ^{2} t \cdot \cos ^{2} t \leq 1 . \frac{1}{\sin ^{2} t} \geq 1$.

Equality is possible only if $\cos ^{4} t+\sin ^{4} t=1$ and $\sin ^{-2} t=1$ simultaneously. This is possible only when $\sin ^{2} t=1$.

Considering that $0<t \leq \frac{\pi}{2}, t=\frac{\pi}{2}$. Then $x=0$.

## Answer: 0

## Recommendations for checking.

The answer $x=0$ is given, but it is not proven that there are no other roots - 0 points.