# How to solve this trigonometric equation: $\sin3x+\sin5x=\sin6x+\sin2x$

Solve for $$x$$ $$\sin3x+\sin5x=\sin6x+\sin2x$$

I tried: $$\sin3x+\sin(2x+3x)=\sin(2*3x)+\sin2x$$ $$\sin3x+\sin2x*\cos3x+\cos2x*\sin3x=2\sin3x*\cos3x+\sin2x$$ ... After that i have no idea what to do. If i try with $$\sin(x+2x)$$ or something similar, it becomes even more scarry.

Hint: use the formula $$\sin x+\sin y = 2\sin {x+y\over 2}\cdot \cos{x-y\over 2}$$ and $$\cos x-\cos y = -2\sin {x+y\over 2}\cdot \sin{x-y\over 2}$$
Rewrite the equation as \begin{align} &\phantom{{}\iff{}}\sin(4x-x)+\sin(4x+x)=\sin(4x+2x)+\sin(4x-2x)\\ &\iff 2\sin 4x\cos x=2\sin 4x\cos 2x\iff\begin{cases}\sin 4x=0\\\cos x=\cos 2x \end{cases} \end{align} Can you proceed from here?