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?
 A: $$\begin{align}\sin(2x)-\sin(x)&=2\sin(x)\cos(x)-\sin(x)\\
&=2\sin(x)(\cos(x)-\frac{1}{2})\tag{1}
\end{align}$$
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:
$$\boxed{0<x<\frac{\pi}{3} \text{ and } \pi<x<\frac{5\pi}{3}}$$
A: 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 $$\dfrac{4m\pi}3\ge0\iff m\ge0, \dfrac{(4m+1)\pi}3\le2\pi\iff m\le\dfrac54\implies m=0,1$$
$$\dfrac{(4n+3)\pi}3\ge0\iff n\ge-\dfrac34;\dfrac{4(n+1)\pi}3\le2\pi\iff n\le\dfrac12\implies n=0$$
A: 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. $x\in(\pi,2\pi-\pi/3)$.

