$x \sinh2y=y \sin2x$

Question

I am working on a problem(in complex analysis) and I am stuck at this point,

I want to solve the following equation: $$x \sinh2y=y \sin2x$$

I can see $x=0$ or $y=0$ satisfy the equation but this deduction was purely through trial and error.

How to rigorously solve the equation? Thanks for reading out, and all the help, and hints.

Answer 1

Hint: $$\sinh(ix)=\frac{e^{ix}-e^{-ix}}{2}=i\sin(x)$$

Here's a different angle of attack:

We know that for $x\geq 0$ and $y\geq0$ $$\sin x \leq x \quad \text{and} \quad \sinh(x)\geq x$$

We deduce that: $$x \sinh 2y \geq 2xy \quad \text{and} \quad y\sin 2x \leq 2xy$$

which means that $$\sinh 2y =2y \quad \text{and} \quad \sin2x=2x$$So that the only real solution in the first quadrant is $y=x=0$. This can be repeated for the remaining quadrants.