## Task B-3.1.

Solve the equation

$$
x^{2}+2 x \sin (x y)+1=0
$$
## First solution.

This equation will have real solutions if the discriminant (of the quadratic equation in $x$) is greater than or equal to 0, that is,

$$
4 \sin ^{2}(x y)-4 \geqslant 0
$$

Since $\sin ^{2}(x y) \leqslant 1$, the above inequality is possible only if $\sin ^{2}(x y)=1$.

Then $\sin (x y)= \pm 1$, that is, $x y= \pm \frac{\pi}{2}+2 k \pi, k \in \mathbb{Z}$.

From $\sin (x y)=1$ and the initial equation, we get $x^{2}+2 x+1=0$, so $x=-1$, and $y=\frac{\pi}{2}+2 k \pi, k \in \mathbb{Z}$. Similarly, for $\sin (x y)=-1$, we get $x=1$, and $y=-\frac{\pi}{2}+2 k \pi, k \in \mathbb{Z}$.

Therefore, the solutions $(x, y)$ of the given equation are $\left(1,-\frac{\pi}{2}+2 k \pi\right)$ and $\left(-1, \frac{\pi}{2}+2 k \pi\right), k \in \mathbb{Z}$.