Question

Solve the inequality: $\sin x>\frac{1}{2}$

Answer 1

Here from the graph, solution of $\sin x>\frac{1}{2}$ is the shaded portion.

Let's solve for

$\sin x=\frac{1}{2}$

or, $\sin x=\sin \frac{\pi }{6}$

or, $x=n\pi +(-1{)}^n\frac{\pi }{6}$

So, $\sin x>\frac{1}{2}$ is in the region.

$(m+1)\pi +(-1{)}^{m+1}\frac{\pi }{6}<x<m\pi +(-1{)}^m\frac{\pi }{6}$

i.e.

$(m\pi +(-1{)}^m\frac{\pi }{6},(m+1)+(-1{)}^m\frac{\pi }{6})$

where $m=0,1,2,3$