Solutions of $\sin2x-\sin x>0$ with $x\in[0,2\pi]$ What are the solutions of this equation with $x\in[0,2\pi]$?
$$\sin2x-\sin x>0$$ 
I took this to
$$(\sin x)(2\cos x-1)>0$$
Now I need both terms to be the same sign. Can you please help me solve this?

Using Prosthaphaeresis Formula,
$$\sin2x-\sin x=2\sin\dfrac x2\cos\dfrac{3x}2$$
Check for $x=0,2\pi$
Else for $0<x<2\pi,\sin\dfrac x2>0$
So we need $$\cos\dfrac{3x}2>0$$  which is possible 
if $2m\pi\le\dfrac{3x}2<2m\pi+\dfrac\pi2\iff\dfrac{4m\pi}3\le x<\dfrac{(4m+1)\pi}3$
or if  $2n\pi+\dfrac{3\pi}2<\dfrac{3x}2\le2n\pi+2\pi\iff\dfrac{(4n+3)\pi}3<x\le\dfrac{4(n+1)\pi}3$
where $m,n$ are arbitrary integers.
Again we need $$\dfrac{4m\pi}3\ge0\iff m\ge0, \dfrac{(4m+1)\pi}3\le2\pi\iff m\le\dfrac54\implies m=0,1$$
$$\dfrac{(4n+3)\pi}3\ge0\iff n\ge-\dfrac34;\dfrac{4(n+1)\pi}3\le2\pi\iff n\le\dfrac12\implies n=0$$
