How do you solve 2sin^2x>=sinx using a sign chart?

Mar 7, 2017

The solution is $\left[\frac{\pi}{6} + 2 k \pi , \frac{5}{6} \pi + 2 k \pi\right] \cup \left[\pi + 2 k \pi , 2 \pi + 2 k \pi\right]$
$k \in \mathbb{Z}$

Explanation:

Let's rearrange and factorise the inequality

$2 {\sin}^{2} x \ge \sin x$

$2 {\sin}^{2} x - \sin x \ge 0$

$\sin x \left(2 \sin x - 1\right) \ge 0$

Let $f \left(x\right) = \sin x \left(2 \sin x - 1\right)$

We need to find the important points for the sign chart

$\sin x = 0$, $\implies$, $x = 0 , \pi , 2 \pi$

$\sin x = \frac{1}{2}$, $\implies$, $x = \frac{\pi}{6} , 5 \frac{\pi}{6}$

Now, we can build the sign chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a a a a a a a}$$0$$\textcolor{w h i t e}{a a a a}$$\frac{\pi}{6}$$\textcolor{w h i t e}{a a a a}$$\frac{5}{6} \pi$$\textcolor{w h i t e}{a a a a a a a a}$$\pi$$\textcolor{w h i t e}{a a a a a a}$$2 \pi$

$\textcolor{w h i t e}{a a a a}$$\sin x$$\textcolor{w h i t e}{a a a a a a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$color(white)(aaa)+$\textcolor{w h i t e}{a a a a a}$$-$

$\textcolor{w h i t e}{a a a a}$$2 \sin x - 1$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$color(white)(aaa)-$\textcolor{w h i t e}{a a a a a}$$-$

$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$color(white)(aaa)-$\textcolor{w h i t e}{a a a a a}$$+$

Therefore,

$f \left(x\right) \ge 0$ when $x \in \left[\frac{\pi}{6} , \frac{5}{6} \pi\right] \cup \left[\pi , 2 \pi\right]$

The solution is $\left[\frac{\pi}{6} + 2 k \pi , \frac{5}{6} \pi + 2 k \pi\right] \cup \left[\pi + 2 k \pi , 2 \pi + 2 k \pi\right]$

$k \in \mathbb{Z}$