<s>[INST] 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 $\sin(x)=-\frac{1}{2}$ ?
 [/INST] You can observe that $-\sin x=\cos(\pi/2+x)$, so your equation becomes
$$
\cos2x=\cos\left(\frac{\pi}{2}+x\right)
$$
which is equivalent to
$$
2x=\frac{\pi}{2}+x+2k\pi
\qquad\text{or}\qquad
2x=-\frac{\pi}{2}-x+2k\pi
$$
and so, after simplifying,
$$
x=\frac{\pi}{2}+2k\pi
\qquad\text{or}\qquad
x=-\frac{\pi}{6}+2k\frac{\pi}{3}
$$
You can now obviate to the “strange” periodicity by dividing the second set of solutions by writing it as
$$
x=-\frac{\pi}{6}+2\cdot3k\frac{\pi}{3}
\quad\text{or}\quad
x=-\frac{\pi}{6}+2\cdot(3k+1)\frac{\pi}{3}
\quad\text{or}\quad
x=-\frac{\pi}{6}+2\cdot(3k+2)\frac{\pi}{3}
$$
(every integer is either a multiple of $3$ or $1$ more than a multiple of $3$ or $2$ more than a multiple of $3$) getting
$$
x=-\frac{\pi}{6}+2k\pi
\quad\text{or}\quad
x=\frac{\pi}{2}+2k\pi
\quad\text{or}\quad
x=\frac{7\pi}{6}+2k\pi
$$
Since these include the other set of solutions, the last sets are the whole sets of solutions.

When dealing with $\sin x=-\frac{1}{2}$, you can remember that $\sin(-x)=-\sin x$, so you can write the equation as
$$
\sin(-x)=\frac{1}{2}
$$
which gives
$$
-x=\frac{\pi}{6}+2k\pi
\qquad\text{or}\qquad
-x=\pi-\frac{\pi}{6}+2k\pi
$$
This becomes
$$
x=-\frac{\pi}{6}-2k\pi
\qquad\text{or}\qquad
x=-\frac{5\pi}{6}-2k\pi
$$
Since $k$ is supposed to be an arbitrary integer, you can also write the last set of solutions as
$$
x=-\frac{\pi}{6}-2(-k-1)\pi
\qquad\text{or}\qquad
x=-\frac{5\pi}{6}-2(-k-1)\pi
$$
so that you get the “principal” solution in the interval $[0,2\pi)$:
$$
x=\frac{11\pi}{6}+2k\pi
\qquad\text{or}\qquad
x=\frac{7\pi}{6}+2k\pi
$$
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