11.15. Find all real solutions of the equation

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

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 $\sin ^{2}(x y) \geqslant \geqslant 1$, 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 $\sin y=-1$.