Product of factors with sign table

Problem

Solve the inequality: $\left(\sin x + \frac{1}{2}\right)\left(2\cos x - \sqrt{3}\right) \ge 0$

Solution

We study the sign of the two factors on $[0,2\pi)$.

• Factor 1: $\sin x + \dfrac{1}{2} \ge 0 \iff \sin x \ge -\dfrac{1}{2}$; it vanishes at $\dfrac{7\pi}{6}, \dfrac{11\pi}{6}$ and is positive on $\left[0,\dfrac{7\pi}{6}\right]\cup\left[\dfrac{11\pi}{6},2\pi\right)$.
• Factor 2: $2\cos x - \sqrt{3} \ge 0 \iff \cos x \ge \dfrac{\sqrt{3}}{2}$; it vanishes at $\dfrac{\pi}{6}, \dfrac{11\pi}{6}$ and is positive on $\left[0,\dfrac{\pi}{6}\right]\cup\left[\dfrac{11\pi}{6},2\pi\right)$.

Sign table:

$x$	$\left[0,\tfrac{\pi}{6}\right)$	$\left(\tfrac{\pi}{6},\tfrac{7\pi}{6}\right)$	$\left(\tfrac{7\pi}{6},\tfrac{11\pi}{6}\right)$	$\left(\tfrac{11\pi}{6},2\pi\right)$
$\sin x+\tfrac{1}{2}$	$+$	$+$	$-$	$+$
$2\cos x-\sqrt{3}$	$+$	$-$	$-$	$+$
product	$+$	$-$	$+$	$+$

The inequality $\ge 0$ is satisfied where the product is positive or zero (including the zeros of the factors): $\boxed{\;x\in\left[0,\frac{\pi}{6}\right]\cup\left[\frac{7\pi}{6},2\pi\right)\;}$ with the usual generalisation by periodicity ($+2k\pi$).

Links

Topics: Trigonometric inequalities
Concepts: Trigonometric inequality · Factorisation · Sign study · Sign table
Functions: Cosine · Sine
Skills: Interpreting a graph · Reasoning by cases · Solving inequalities
Exercise type: Solving an inequality