## Task B-4.1.

Real numbers $x, y$ are solutions to the system of equations

$$
\begin{gathered}
2 \sqrt{2} x \cos t + 3 y \sin t = 6 \sqrt{2} \\
2 \sqrt{2} x \sin t - 3 y \cos t = 0
\end{gathered}
$$

For which values of the parameter $t \in (0, \pi)$ is the product $x y$ equal to 3?
## Solution.

$$
\begin{aligned}
2 \sqrt{2} x \cos t+3 y \sin t & =6 \sqrt{2} / \cdot \cos t \\
2 \sqrt{2} x \sin t-3 y \cos t & =0 / \cdot \sin t \\
\hline 2 \sqrt{2} x \cos ^{2} t+3 y \sin t \cos t & =6 \sqrt{2} \cos t \\
2 \sqrt{2} x \sin ^{2} t-3 y \cos t \sin t & =0 .
\end{aligned}
$$

By adding these two equations, we get $x=3 \cos t$.

1 point

Notice that $\cos t \neq 0$, otherwise $x=0$ and $x \cdot y \neq 3$.

1 point

Then from the second equation, we have:

$$
\begin{gathered}
6 \sqrt{2} \cos t \sin t-3 y \cos t=0 /: 3 \cos t \\
y=2 \sqrt{2} \sin t \\
x \cdot y=3 \Rightarrow 3 \cos t \cdot 2 \sqrt{2} \sin t=3 \\
\Rightarrow 3 \sqrt{2} \sin 2 t=3 \Rightarrow \sin 2 t=\frac{\sqrt{2}}{2}
\end{gathered}
$$

Thus, $2 t=\frac{\pi}{4}$ or $2 t=\frac{3 \pi}{4}$, i.e., $t \in\left\{\frac{\pi}{8}, \frac{3 \pi}{8}\right\}$.