Unable to reach the desired answer in trigonometry. The question is: 
If $\sin x + \sin y = \sqrt3 (\cos y - \cos x)$
show that $\sin 3x + \sin 3y= 0 $
This is what I have tried: 


*

*Squaring of the first equation (Result: Failure)

*Tried to use the $\sin(3x)$ identity but got stuck in the middle steps because I couldn't simplify it any further.


Can someone provide any hint/ suggestion? 

I was trying to find out how the condition was conceived.
$$\sin3x+\sin3y=0\implies\sin3x=\sin(-3y)$$
$\implies3x=180^\circ n+(-1)^n(-3y)$ where $n$ is any integer
$\iff x=60^\circ n+(-1)^{n+1}y$
If $n$ is even $=2m$(say), $x=120^\circ m-y$
$$\implies x+y\equiv\begin{cases}0 &\mbox{if }3\mid m\\120^\circ& \mbox{if } n \equiv1\pmod3\\240^\circ& \mbox{if } n\equiv2\pmod3 \end{cases}\pmod{360^\circ}$$
$\implies\sin\dfrac{x+y}2=\tan\dfrac{x+y}2=0$ or $\tan\dfrac{x+y}2=\pm\sqrt3$
Similarly for odd $n=2m+1$(say),
$\cos\dfrac{x-y}2=\cot\dfrac{x-y}2=0$ or $\cot\dfrac{x-y}2=\pm\sqrt3$
Here the condition chosen $$0=\sin\dfrac{x+y}2\left(\cot\dfrac{x-y}2-\sqrt3\right)=\dfrac{2\sin\dfrac{x+y}2\cos\dfrac{x-y}2-\sqrt3\cdot2\sin\dfrac{x+y}2\sin\dfrac{x-y}2}{2\sin\dfrac{x-y}2}=\dfrac{\sin x+\sin y-\sqrt3(\cos y-\cos x)}{2\sin\dfrac{x-y}2}$$ with $\sin\dfrac{x-y}2\ne0$ as  $\cot\dfrac{x-y}2=\sqrt3$
which could easily be $$\sin\dfrac{x+y}2\left(\cot\dfrac{x-y}2+\sqrt3\right)=0\iff\sin x+\sin y=-\sqrt3(\cos y-\cos x)$$
Or $$\cos\dfrac{x-y}2\left(\tan\dfrac{x+y}2\pm\sqrt3\right)=0\iff\sin x+\sin y=\pm\sqrt3(\cos y+\cos x)$$
