Slovenia — National Math Olympiad, 2012

Denote $y = \pi \sin^2 x$. Because $\sin(\pi \cos^2 x) = \sin(\pi(1 - \sin^2 x)) = \sin(\pi - \pi \sin^2 x) = \sin(\pi \sin^2 x)$, we get the equation $\cos y + \sin y = 1$.

Because $\cos y + \sin y = \sin y + \sin\left(\frac{\pi}{2} - y\right) = 2\sin\frac{\pi}{4}\cos\frac{2y-\frac{\pi}{2}}{2} = 1$, we get $\cos\left(y - \frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$, and hence $y = \frac{\pi}{4} \pm \frac{\pi}{4} + 2k\pi = \pi \sin^2 x \in [0, \pi]$, it must be $k=0$. Thus $y=0$ or $y=\frac{\pi}{2}$.

If $y=0$, we get $x = n\pi$ where $n$ is an integer.

For $y = \frac{\pi}{2}$ it must hold $\sin^2 x = \frac{1}{2}$, from which we derive $\sin x = \pm\frac{\sqrt{2}}{2}$ or $x = \frac{\pi}{4} + n\frac{\pi}{2}$ where $n$ is an integer.