3. For which real numbers $x$ and $y$ does the equality

$$
4 x^{2}+12 x \sin (x y)+9=0 ?
$$
Solution. We have

$$
\begin{aligned}
4 x^{2}+12 x \sin (x y)+9 & =4 x^{2}+12 x \sin (x y)+9 \sin ^{2}(x y)-9 \sin ^{2}(x y)+9 \\
& =(2 x+3 \sin (x y))+9 \cos ^{2}(x y)
\end{aligned}
$$

we rewrite the given equation in the form

$$
(2 x+3 \sin (x y))+9 \cos ^{2}(x y)=0
$$

From the last equation, it follows that $2 x+3 \sin (x y)=0$ and $\cos (x y)=0$. From the second equation, we get: $\quad x y=\frac{\pi}{2}+k \pi, k \in \mathbb{Z}$, then substituting into the first equation we will have $x_{1}=-\frac{3}{2}, x_{2}=\frac{3}{2}$, and accordingly $y_{1}=-\frac{\pi}{3}+\frac{2 k \pi}{3}, y_{2}=\frac{\pi}{3}+\frac{2 k \pi}{3}, k \in \mathbb{Z}$.