3.1. Find all roots of the equation $\sin (\pi \cos 2 x)=\cos \left(\pi \sin ^{2} x\right)$, lying in the interval $\left[-\frac{5 \pi}{3} ;-\frac{5 \pi}{6}\right]$. In the answer, record the sum of these roots divided by $\pi$ (in radians), rounding it to two decimal places if necessary.
Answer: $-6.25$.

Solution. Let $t=\pi \sin ^{2} x$. Then $\pi \cos 2 x=\pi-2 t$, so the equation takes the form $\sin (\pi-2 t)=\cos t \Leftrightarrow \sin 2 t=\cos t \Leftrightarrow(2 \sin t-1) \cos t=0 \Leftrightarrow\left[\begin{array}{l}t=\frac{\pi}{2}+\pi k, \\ t=(-1)^{l} \frac{\pi}{6}+\pi l, k, l \in \mathbb{Z} .\end{array}\right.$

Since $0 \leqslant t \leqslant \pi$, these equalities are possible only for $k=0, l=0$ and $l=1$. Therefore, $\sin ^{2} x=\frac{1}{2}, \frac{1}{6}$ or $\frac{5}{6}$. Considering that $\arcsin \sqrt{\frac{5}{6}}=\frac{\pi}{2}-\arcsin \frac{1}{\sqrt{6}}$, we obtain the solution to the original equation: $x=\frac{\pi}{4}+\frac{\pi n}{2}, x= \pm \arcsin \frac{1}{\sqrt{6}}+\frac{\pi m}{2}, n, m \in \mathbb{Z}$. Since $\alpha=\arcsin \frac{1}{\sqrt{6}}<\frac{\pi}{6}\left(=\arcsin \frac{1}{2}\right)$, on the interval $\left[-\frac{5 \pi}{3} ;-\frac{5 \pi}{6}\right]$ there are 5 roots $-\pi \pm \alpha,-\frac{5 \pi}{4},-\frac{3 \pi}{2} \pm \alpha$, the sum of which is $-\frac{25 \pi}{4}$.