Solving $\cos(2x)-\sin(x)= 0$ within the domain $[0,2\pi]$. Why am I missing some solutions? I‘m trying to figure this one out:

Solve for $x$ within the domain $[0,2\pi]$:
  $$\cos(2x)-\sin(x)= 0$$

I figured this is $\sin(x)=-1$ and $\sin(x) = \frac 12$, which, when taking the $\arcsin$, gives $x=\frac{\pi}6$ and $x=-\frac{\pi}2$.
Now my question is: 

I‘m supposed to get $\dfrac{\pi}6$, $\dfrac{5\pi}6$, $\dfrac{3\pi}2$, but adding $2\pi$ gets me only $\dfrac{\pi}6$, $\dfrac{3\pi}2$, $\dfrac{7\pi}2$ within the $[0,2\pi]$ domain.

Any idea what I‘m doing wrong here?
 A: \begin{align*}
\cos(2x) - \sin x & = 0\\
1 - 2\sin^2x - \sin x & = 0\\
1 - \sin x - 2\sin^2x & = 0\\
1 - 2\sin x + \sin x - 2\sin^2x & = 0\\
1(1 - 2\sin x) + \sin x(1 - 2\sin x) & = 0\\
(1 + \sin x)(1 - 2\sin x) & = 0
\end{align*}
\begin{align*}
1 + \sin x & = 0 &  1 - 2\sin x & = 0\\
\sin x & = -1 & -2\sin x & = -1\\
& & \sin x & = \frac{1}{2}
\end{align*}
Consider the diagram below.

The sine of an angle in standard position (vertex at the origin and initial side on the positive $x$-axis) is the $y$-coordinate of the point where the terminal side of the angle intersects the unit circle.  Two angles have the same sine if their terminal sides intersect the unit circle at points with the same $y$-coordinate.  Such angles are symmetric with respect to the $y$-axis. Thus, $\sin\theta = \sin\varphi$ if $\varphi = \pi - \theta$.  Any two coterminal angles will also have the same sine.  Hence, $\sin\theta = \sin\varphi$ implies 
$$\varphi = \theta + 2k\pi, k \in \mathbb{Z}$$
or 
$$\varphi = \pi - \theta + 2k\pi, k \in \mathbb{Z}$$ 
You found that a particular solution of the equation $\sin x = -1$ is 
$$x = \arcsin(-1) = -\frac{\pi}{2}$$
Hence, all solutions of that equation have the form
\begin{align*}
\theta & = -\frac{\pi}{2} + 2k\pi, k \in \mathbb{Z} & \theta &  = \pi - \left(-\frac{\pi}{2}\right) + 2k\pi\\
       & & & = \frac{3\pi}{2} + 2k\pi
\end{align*}
If you list the solutions, you will notice that both equations yield the same set of solutions.  The only one within the interval $[0, 2\pi]$ is $\dfrac{3\pi}{2}$.  
You found that one solution of the equation $\sin x = \dfrac{1}{2}$ is $$x = \arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6}$$  Hence,  the general solution is
\begin{align*}
x & = \frac{\pi}{6} + 2k\pi, k \in \mathbb{Z} & x & = \pi - \frac{\pi}{6} + 2k\pi, k \in \mathbb{Z}\\
  & & & = \frac{5\pi}{6} + 2k\pi, k \in \mathbb{Z}
\end{align*}
Of these angles, the only ones in the interval $[0, 2\pi]$ are $\dfrac{\pi}{6}$ and $\dfrac{5\pi}{6}$. 
By only considering the arcsine of the angle, you missed the angle with the same sine that is obtained by reflection in the $y$-axis.  There is only one such angle within each period if the terminal side of the angle lies on the $y$-axis (as is the case for $\theta = -\frac{\pi}{2}$) and two such angles otherwise (as is the case for $\theta = \frac{\pi}{6}$). 
A: You are correct to add $2 \pi$ to $-\frac{\pi}{2}$ to get the angle $\frac{3\pi}{2} \in [0,2 \pi]$.
You shouldn't add $2\pi$ to $\frac{\pi}{6}$; that's already an angle in $[0,2\pi].$  
Observe that the angle $\frac{5\pi}{6},$ the reflection of $\frac{\pi}{6}$ in the $y-$axis, also gives the same sine value.  (Look at the unit circle: the points corresponding to these angles have the same $y$-value.)
Whereas, $\frac{3\pi}{2}$ corresponds to a $y$-value of $-1$, and there is no other angle in $[0,2\pi]$ which does.
