## Task A-3.3.

Determine all real numbers $x$ for which there exists a real number $y$ such that

$$
\frac{\sin (2 x)}{\sin (2 y)}=4 \sin (x+y) \cos (x-y)
$$
## Solution.

Notice that $\sin (2 y) \neq 0$ must hold for the equation to be well-defined.

By applying the product-to-sum formula, the equation becomes

$$
\frac{\sin (2 x)}{\sin (2 y)}=2(\sin (2 x)+\sin (2 y))
$$

or equivalently,

$$
\sin ^{2}(2 y)+\sin (2 x) \sin (2 y)-\frac{\sin (2 x)}{2}=0
$$

From here, we see that $\sin (2 y)=0$ if and only if $\sin (2 x)=0$. Therefore, it is necessary that $\sin (2 x) \neq 0$.

Solving the quadratic equation for $\sin (2 y)$, we get

$$
\sin (2 y)=\frac{-\sin (2 x) \pm \sqrt{\sin ^{2}(2 x)+2 \sin (2 x)}}{2}
$$

Let $t:=\sin (2 x)$. At least one of the equations

$$
\sin (2 y)=\frac{-t+\sqrt{t^{2}+2 t}}{2} \quad \text { and } \quad \sin (2 y)=\frac{-t-\sqrt{t^{2}+2 t}}{2}
$$

has a solution $y \in \mathbb{R}$ if $t^{2}+2 t \geqslant 0$, and if the expression on the right-hand side of the corresponding equation lies in the interval $[-1,1]$.

Since $t=\sin (2 x) \in[-1,1]$, we have $2+t>0$, so

$$
t^{2}+2 t=t(t+2) \geqslant 0
$$

if and only if $t \geqslant 0$, which together with the condition $\sin (2 x) \neq 0$ gives $t>0$.

In this case,

$$
\frac{-t+\sqrt{t^{2}+2 t}}{2} \geqslant \frac{-t+\sqrt{t^{2}}}{2}=0>-1
$$

and

$$
\frac{-t+\sqrt{t^{2}+2 t}}{2} \leqslant \frac{-t+\sqrt{t^{2}+2 t+1}}{2}=\frac{-t+(t+1)}{2}=\frac{1}{2}<1
$$

The condition $\sin (2 x)>0$ is equivalent to

$$
x \in \bigcup_{k \in \mathbb{Z}}\left\langle k \pi, \frac{(2 k+1) \pi}{2}\right\rangle
$$

Thus, these are all real numbers $x$ with the desired property.

1 point