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
 A: HINT
Cosine is the horizontal coordinate, so $\cos x < 0$ in II and III quadrants. Note that
$$
\cos(180^\circ - \theta) = -\cos \theta = \cos (180^\circ + \theta)
$$
A: $$\sin(2x)+\sin(x)=0 $$
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 $
A: We know that $\cos 60^\circ = \frac{1}{2}$.  
The cosine of an angle is defined to be the $x$-coordinate of the point where the terminal side of an angle in standard position (initial side on the positive $x$-axis and vertex at the origin) intersects the unit circle.  Therefore, the cosine function is positive if the terminal side of the angle lies in the first quadrant, fourth quadrant, or on the positive $x$-axis; $0$ if the terminal side of the angle lies on the $y$-axis; and negative if the terminal side of the angle lies in the second quadrant, third quadrant, or on the negative $x$-axis.
Now consider the diagram below.

Observe that $\cos(\pi - \theta) = \cos(180^\circ - \theta) = -\cos\theta$.  Hence, 
$$\cos(180^\circ - 60^\circ) = \cos(120^\circ) = -\cos(60^\circ) = -\frac{1}{2}$$
Also, observe that $\cos(\pi + \theta) = \cos(180^\circ - \theta) = -\cos\theta)$.  Hence,
$$\cos(180^\circ + 60^\circ) = \cos(240^\circ) = -\cos(60^\circ) = -\frac{1}{2}$$
A: $\cos(x) = -1/2$ in the quadrants where $cos(x)$ is negative, so quadrants 2 and 3. You are correct about $\sin(x)$. Look at $\cos^{-1}(x)$ and see where it is $1/2$.
A: If $\sin x=0$, then $x=k\pi$, where $k\in\mathbb{Z}$. If $\cos x=-\frac{1}{2}$, then either $x=\frac{2\pi}{3}+2k\pi, k\in\mathbb{Z}$ or $\cos x=\frac{4\pi}{3}+k\pi, k\in\mathbb{Z}.$
