2. Find all real solutions of the equation $x^{2}+2 x \sin (x y)+1=0$.
Solution If we consider the given equation as a quadratic equation in terms of $\mathrm{x}$, then its discriminant will be equal to $4 \sin ^{2}(\mathrm{X}) - 1$. The discriminant must be non-negative, so $\sin ^{2}(\mathrm{X}) \geq 1$, i.e., $\sin (\mathrm{X})= \pm 1$. Let $\sin (\mathrm{x})=1$. Then we get the equation $x^{2}+2 x+1=0$, solving which we find $x=-1$. Therefore, $\sin (-y)=1$, i.e., $y=-\pi / 2+\pi k$, $\mathrm{k} \in \mathrm{Z}$. Let $\sin (\mathrm{Xy})=-1$. Then we get the equation $\mathrm{x}^{2}-2 \mathrm{x}+1=0$, solving which we find $\mathrm{X}$ $=1$. Therefore, $\sin (y)=-1$, i.e., $y=-\pi / 2+\pi k, k \in Z$.

Answer: $x= \pm 1, y=-\pi / 2+\pi k, k \in Z$