How do I prove when $2\sin^2x$ is greater than $\sin x $? I need to show when 

$2\sin^2x \geq \sin x $ for $x$ between $0$ and $2\pi$. 

I see that for values $\pi/6$ and $5\pi/6$ and after $\pi$. How do I prove or reason out these answers?
If $x\geqslant\pi$, then $2\sin^2x\geqslant0\geqslant\sin x$. If $x=0$ this is again true.

If $0&lt;x&lt;\pi$, then $\sin x&gt;0$ and therefore$$2\sin^2x\geqslant\sin x\iff 2\sin x\geqslant 1\iff \sin x\geqslant\frac12$$and this takes place if and only of $\frac\pi6\leqslant x\leqslant\frac{5\pi}6$.
$2\sin^2x \geq \sin x $ can simplify to $2\sin x \geq 1$ after factoring out a $\sin x$.

$2\sin x \geq 1$ is true only when $\sin x \geq 1/2$. 

Therefore $30^\circ \leq x \leq 150^\circ$. $(x \neq 0,360)$