solve $\sin 2x + \sin x = 0 $ using addition formula 
$\sin 2x + \sin x = 0 $

Using the addition formula, I know that
$\sin 2x = 2\sin x \cos x$
=> $2\sin x \cos x + \sin x = 0$
=> $\sin x(2\cos x + 1) = 0$
=> $\sin x = 0$ and $\cos x = -\frac{1}2 $
I know that $\sin x = 0$ in first and second quadrant so $x = 0$ and $x = 180$
What I do not know is what to do with $\cos x = -\frac{1}2$ and which quadrants this applies to.
The book I got the question from gives the following answer which does not make sense to me:
0, 120, 180, 240, 360

$$\sin(2x)+\sin(x)=0 $$
First, you have to remember that: 
$$\sin(a)+\sin(b)=2\sin(\frac{a+b}{2})\cos(\frac{a-b}{2})$$

Aplying this in the equation:
$$2\sin(\frac{3x}{2})\cos(\frac{x}{2})=0$$
$$\sin(\frac{3x}{2})\cos(\frac{x}{2})=0$$
Finally, you have two equations: 
$$\sin(\frac{3x}{2})=0$$
$$\cos(\frac{x}{2})=0$$

For the first, the solution are $x=\frac{2n\pi}{3} / n=0,1,2 $
And for the second are $x=n\pi / n=1,3 $
