1. Solve the equation $x^{2}-2 x \cdot \sin (x \cdot y)+1=0$.
Solution. We will interpret the left side of the equation as a quadratic trinomial in terms of $x$. For the roots to exist, the discriminant must be non-negative, i.e., $D=4 \sin ^{2}(x \cdot y)-4 \geq 0 \Leftrightarrow \sin ^{2}(x \cdot y)=1 \Leftrightarrow \cos 2 x y=-1 \Leftrightarrow x y=\frac{\pi}{2}+\pi n, n \in \mathbb{Z}$. For even $n$, we get the equation $x^{2}-2 x+1=0 \Leftrightarrow x=1$, and for odd $n$, we find $x=-1$. The left side of the equation is an even function of $x$, so for $x=1$ and $x=-1$, the corresponding values of $y$ will be the same.

Answer: $\left( \pm 1, \frac{\pi}{2}+2 \pi k\right), k \in \mathbb{Z}$.