Is $a=b$ the only possibility? 
In the following: $$2a-2b+(b^2-a^2)\operatorname{sinh} (by)\cos(ax)=0, \forall x,y \in \mathbb{R},$$ does it result automatically that $a=b$ in order for this to hold? 

Is there any other possibility? Why?

Say $a\ne b$ then we have $$2-(b+a)\sinh(by)\cos(ax)=0$$ for each $x,y$. So for $y=0$ we get $2=0$.
A contradiction.
