Task B-3.6. Solve the system of equations on the interval $\left\langle 0, \frac{\pi}{2}\right\rangle$

$$
\left\{\begin{array}{l}
\frac{\cos x}{\cos y}=2 \cos ^{2} y \\
\frac{\sin x}{\sin y}=2 \sin ^{2} y
\end{array}\right.
$$
## Solution.

By adding the given equations, we have

$$
\frac{\cos x}{\cos y}+\frac{\sin x}{\sin y}=2
$$

After multiplying by $\sin y \cos y$, we get

$$
\begin{aligned}
& \sin x \cos y+\cos x \sin y=2 \sin y \cos y \\
& \sin (x+y)=\sin 2 y
\end{aligned}
$$

We have two cases:

First case. From $x+y=2 y$, it follows that $x=y$, so we have

$$
\begin{aligned}
& \frac{\sin x}{\sin x}=2 \sin ^{2} x \\
& 2 \sin ^{2} x=1 \\
& \sin ^{2} x=\frac{1}{2}
\end{aligned}
$$

The equation $\sin x= \pm \frac{\sqrt{2}}{2}$ on $\left\langle 0, \frac{\pi}{2}\right\rangle$ has the solution $x_{1}=\frac{\pi}{4}$.

One solution to the system is $x=\frac{\pi}{4}, y=\frac{\pi}{4}$.

Second case. From $x+y=\pi-2 y$, it follows that $x=\pi-3 y$, so we have

$$
\begin{aligned}
& \frac{\sin (\pi-3 y)}{\sin y}=2 \sin ^{2} y \\
& \frac{\sin 3 y}{\sin y}=2 \sin ^{2} y \\
& \frac{3 \sin y-4 \sin ^{3} y}{\sin y}=2 \sin ^{2} y \\
& 3-4 \sin ^{2} y=2 \sin ^{2} y \\
& 6 \sin ^{2} y=3 \\
& 2 \sin ^{2} y=1 \\
& \sin ^{2} y=\frac{1}{2}
\end{aligned}
$$

The equation $\sin y= \pm \frac{\sqrt{2}}{2}$ on $\left\langle 0, \frac{\pi}{2}\right\rangle$ has the solution $y_{1}=\frac{\pi}{4}$.

This again gives us the same solution to the system.

## Second Solution.

$\mathrm{In}$ the same way, we reduce the system to the equation

$$
\sin (x+y)=\sin 2 y
$$

We can solve this equation using transformation formulas.

$$
\begin{aligned}
& \sin (x+y)-\sin 2 y=0 \\
& 2 \cos \frac{x+3 y}{2} \sin \frac{x-y}{2}=0
\end{aligned}
$$

Then we have

$$
\cos \frac{x+3 y}{2}=0 \quad \text { or } \quad \sin \frac{x-y}{2}=0
$$

This implies

$$
\frac{x+3 y}{2}=\frac{\pi}{2} \quad \text { or } \quad \frac{x-y}{2}=0
$$

From which we get $x=\pi-3 y$ or $x=y$.

The rest proceeds analogously to the first solution.