3-ча 1. Find all real solutions of the equation

\[
x^{2}+2 x \sin (x y)+1=0
\]
Solve 1. Answer: \(x= \pm 1, y=-\frac{\pi}{2}+k \pi\).

If we consider the given equation as a quadratic equation in terms of \(x\), then its discriminant will be equal to \(4\left(\sin ^{2}(x y)-1\right)\). The discriminant must be non-negative, so \(\sin ^{2}(x y) \geqslant 1\), i.e., \(\sin (x y)= \pm 1\). The solutions of the equation \(x^{2} \pm 2 x+1=0\) are \(x=\mp 1\). Further, if \(\sin (\mp y)= \pm 1\), then \(\sin y=-1\).