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?

When $\sin x>0$, i.e. $x\in(0,\pi)$, the inequation reduces to $\cos x>1/2$, i.e. $x\in(0,\pi/3)$.
When $\sin x<0$, i.e. $x\in(\pi,2\pi)$, the inequation reduces to $\cos x<1/2$, i.e. $x\in(\pi,2\pi-\pi/3)$.

