Showing whether two numbers are equal or not $\dfrac{\sin (2x+y)}{\sin (2x)} =\dfrac{\sin (x+2y)}{\sin (2y)}$,where $0<x,y\le\dfrac{\pi}{4}$ .
Can I show that $x=y $ or find two numbers $x,y$ such that $x\not=y$?

Edited: I think I got it...there was typo in the first try.
After cross-multiplication, we get
$[2\sin y\sin(2x+y)]\cos y-[2\sin x\sin(x+2y)]\cos x=0$
$\Rightarrow[\cos(2x)-\cos2(x+y)]\cos y-[\cos(2y)-\cos2(x+y)]\cos x=0$
$\Rightarrow\cos2(x+y)[\cos x-\cos y]+[(2\cos^2x-1)\cos y-(2\cos^2y-1)\cos x]=0$
$\Rightarrow[\cos x-\cos y][\cos2(x+y)+2\cos x\cos y+1]=0$
$\Rightarrow[\cos x-\cos y][\cos^2(x+y)+\cos x\cos y]=0$
Note that in the specified range, the second factor is strictly positive. Hence we must have $\cos x=\cos y\Rightarrow x=y$
