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?
 A: If $x\geqslant\pi$, then $2\sin^2x\geqslant0\geqslant\sin x$. If $x=0$ this is again true.
If $0<x<\pi$, then $\sin x>0$ 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 $\frac\pi6\leqslant x\leqslant\frac{5\pi}6$.
A: Consider $f(x) := 2\sin^2(x) - \sin(x)$. Essentially, your quesion boils down to finding where $f(x) \ge 0$. We will first find where $f(x)$ is zero. As you have already observed,
$$f(x) = 0 \Rightarrow 2\sin^2(x) - \sin(x) = 0 \Rightarrow (2\sin(x) - 1)\sin(x) = 0$$
Which means that $x = 0, \pi/6, 5\pi/6, \pi$, and $2\pi$ are our "critical points". Since these are the only points where $f$ is zero, the continuity of $f$ implies that it must be strictly positive or negative on the intervals $(0, \pi/6)$, $(\pi/6, 5\pi/6)$, $(5\pi/6, \pi)$, and $(\pi, 2\pi)$. Simply choose one "test point" inside these intervals, and you will know immediately whether each respective interval is positive or negative.
You'll find that both $[\pi/6, 5\pi/6]$ and $[\pi, 2\pi]$ are positive (we have closed brackets here because we allow $f$ to be equal to zero as well). Hence, your inequality is satisfied on $[\pi/6, 5\pi/6]$ and $[\pi, 2\pi]$.
A: Hint: If $\sin x\leq 0$, the LHS is nonnegative and the RHS is negative. Otherwise, we can divide each side by $\sin x$...
A: Case 1. If $\sin x =0$ you know the values.
Case 2. If $\sin x \gt 0$ then you can divide through by $2\sin x$ and you need $\sin x \ge \frac 12$ (and if this is true you have $\sin x \gt 0$ as well).
Case 3. If $\sin x \lt 0$ when you divide through by $2 \sin x$ the inequality swaps and you get $\frac 12 \ge \sin x$ and since you also need $\sin x \lt 0$ this is the condition - you can combine this with the first case to give $\sin x \le 0$.
Now use your knowledge of the sine function to identify the parts this picks out.
A: I'll just combine everyone's answer and comment here.
Let $y=\sin \theta$. We have $f(y) = 2y^2 - y = y(2y-1)$. The function $f(y)\ge 0$ everywhere except when $0 < y < \frac12$.
On the classical trig unit circle, the angle $\theta$ that gives $y$ values between $0$ and $\frac12$ lies in $(0, \frac\pi6) \cup (\frac{5\pi}{6}, \pi)$.
So the inequality only holds when $x \in [\frac\pi6, \frac{5\pi}{6}] \cup [\pi, 2\pi].$
A: $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 $30^\circ \leq x \leq 150^\circ$. $(x \neq 0,360)$
