Elementary inequality with the sine

Problem

Solve the inequality: $\sin x \ge \frac{\sqrt{3}}{2}$

Solution

On the unit circle $\sin x = \dfrac{\sqrt{3}}{2}$ for $x=\dfrac{\pi}{3}$ and $x=\dfrac{2\pi}{3}$. The inequality $\sin x \ge \dfrac{\sqrt{3}}{2}$ requires the points above the horizontal line $y=\dfrac{\sqrt{3}}{2}$, that is the arc between the two values. On $[0,2\pi)$: $x\in\left[\frac{\pi}{3}, \frac{2\pi}{3}\right]$ Generalising with periodicity: $\boxed{\;x\in\left[\frac{\pi}{3}+2k\pi,\ \frac{2\pi}{3}+2k\pi\right]\;}$

Links

Topics: Trigonometric inequalities
Concepts: Trigonometric inequality · Trigonometric inequality reading the circle
Functions: Sine
Skills: Interpreting a graph · Solving inequalities
Exercise type: Solving an inequality