How to solve this trigonometric equalities? We have to solve $ \tan(x) = 2 \sin(x)$ on $0 \leq x \leq 2 \pi$.
I rewrote it to $\sin(x) = \sin(2x)$ using $2\sin(x)\cos(x) = \sin(2x)$. 
But how can I now find the set of solutions by hand? 

Or you can just substitute $tan(x)=\frac{sin(x)}{cos(x)}$ and then we have 
$\frac{sin(x)}{cos(x)}=2sin(x)\rightarrow sin(x)(\frac{1}{cos(x)}-2)=0$
Then $sin(x)=0$ or $cos(x)=\frac{1}{2}$
$sin(x)=0$ when $x=0$ or $x=\pi$ or $x=2\pi$ we have 3 solutions that belong to $[0,2\pi]$
and 
$cos(x)=\frac{1}{2}\rightarrow cos(x)=cos(\frac{\pi}{3})$
$x=2k\pi+\frac{\pi}{3}$ or $x=2k\pi-\frac{\pi}{3}$ So we have 2 more solutions $\frac{\pi}{3}$ and $2\pi-\frac{\pi}{3}$
