TITLE: Solve $\sin2x +\sin x = 0$ algebraically
QUESTION [5 upvotes]: I am studying for a final and came across a review question that I have no idea how to do. The question is "Solve the equation $\sin(2x) + \sin(x) = 0$ on the interval $[0, 2\pi)$. 
I can graph it and get the solutions $0, 2\pi/3, \pi$, and $4\pi/3$ where it crosses zero between $0$ and $2\pi$. My question is how do you do it algebraically?

REPLY [4 votes]: $$\sin2x=-\sin x=\sin(-x)$$
$$\implies2x=n\pi+(-1)^n(-x)$$ where $n$ is any integer
If $n$ is even, $=2m$(say)  $$2x=2m\pi-x\iff x=\frac{2m\pi}3$$
If $n$ is odd, $=2m+1$(say)   $$2x=(2m+1)\pi+x\iff x=(2m+1)\pi$$