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?

$$\begin{align}\sin(2x)-\sin(x)&=2\sin(x)\cos(x)-\sin(x)\\
&=2\sin(x)(\cos(x)-\frac{1}{2})\tag{1}
\end{align}$$
We examine the two factors:
$$\cos(x)=\frac{1}{2}\Rightarrow x=\frac{\pi}{3},\frac{5\pi}{3}$$
And $$\cos(x)-\frac{1}{2}>0,\, 0<x<\frac{\pi}{3} \text{ and }  \frac{5\pi}{3}<x<2\pi$$
$$\cos(x)-\frac{1}{2}<0,\, \frac{\pi}{3}<x<\frac{5\pi}{3}$$
The other term is $\sin(x)$: 
$$\sin(x)>0,\, 0<x<\pi$$
$$\sin(x)<0,\, \pi<x<2\pi$$
So in the following interval the product is positive:
$$\boxed{0<x<\frac{\pi}{3} \text{ and } \pi<x<\frac{5\pi}{3}}$$
