4A. Determine $x, y \in\left(0, \frac{\pi}{2}\right)$ for which the equalities hold

$$
\cos x=2 \cos ^{3} y, \sin x=2 \sin ^{3} y
$$
Solution. For $\sin y, \cos y \neq 0$ we obtain $\frac{\sin x}{\sin y}=2 \sin ^{2} y$ and $\frac{\cos x}{\cos y}=2 \cos ^{2} y$ from which it follows that $\frac{\sin x}{\sin y}+\frac{\cos x}{\cos y}=2, \sin (x+y)=\sin 2 y$, so $2 \sin \frac{x-y}{2} \cos \frac{x+3 y}{2}=0$. Given $x, y \in\left(0, \frac{\pi}{2}\right)$, we have $\frac{x-y}{2} \in\left(-\frac{\pi}{4}, \frac{\pi}{4}\right)$ and for $\sin \frac{x-y}{2}=0$ it must be that $x=y$. Substituting into the initial conditions, we get $\sin x=2 \sin ^{3} x$ or $\sin x=\frac{1}{\sqrt{2}}$, i.e., $x=y=\frac{\pi}{4}$. On the other hand, given $x, y \in\left(0, \frac{\pi}{2}\right)$, we have $\frac{x+3 y}{2} \in(0, \pi)$ and for $\cos \frac{x+3 y}{2}=0$ it must be that $x+3 y=\pi$. Substituting into the initial conditions, we get $\sin 3 y=\sin (\pi-3 y)=\sin x=2 \sin ^{3} y$ from which $3 \sin y-4 \sin ^{3} y=2 \sin ^{3} y$, i.e., $\sin y=2 \sin ^{3} y$, or $\sin y=\frac{1}{\sqrt{2}}$ and $y=\frac{\pi}{4}$, and therefore $x=\pi-3 y=\frac{\pi}{4}$.

## Fourth Year