Solve $\sin(x)=-\frac{1}{2}$ I have to solve the following equation for $x$

$$\cos(2x)+\sin(x)=0$$

After simplification i got 

$\sin(x)=1$ or $\sin(x)=-\frac{1}{2}$

$\Rightarrow x=90^0$

But don't know how to solve for $x$ for $\sin(x)=-\frac{1}{2}$ ?
If $\sin x=-1/2$, then $x=210^{\circ}$ or $x=330^{\circ}$ are solutions. In the third and fourth quadrants sine becomes negative. Alternatively you can say $x=-30^{\circ}$ or $x=-150^{\circ}$.

The trick here is to have a mental picture of what the sine function is doing. Also the values are the same when you add or subtract $360^{\circ}$.

The way I was taught to remember which one is positive was remembering a mnemonic for the letters A, S, T and C.

A = All

S = sine

T = tangent

C = cosine
If $\space sin(x)=-1/2\space $ then $\space x=\pi n+\frac{\pi}{6}\cdot(-1)^{n+1}$

Edit: $\space n\in Z$