Product of factors and sign table

Let us see the sign-study method applied to an inequality already written as a product of two factors: it is enough to study the sign of each one separately and combine them in a single table.

Example — Teacher's handout, 4C 2017-18

Solve $\left(\sin x - \dfrac{1}{2}\right)\left(\dfrac{\sqrt{2}}{2} - \cos x\right) \ge 0$.

Factor 1: $\sin x - \dfrac{1}{2} \ge 0 \iff \sin x \ge \dfrac{1}{2}$. On the unit circle, $\sin x = \dfrac{1}{2}$ for $x=\dfrac{\pi}{6}$ and $x=\dfrac{5\pi}{6}$; it is positive above the line $y=\dfrac{1}{2}$, that is on $\left[\dfrac{\pi}{6}, \dfrac{5\pi}{6}\right]$.

Factor 2: $\dfrac{\sqrt{2}}{2} - \cos x \ge 0 \iff \cos x \le \dfrac{\sqrt{2}}{2}$. We have $\cos x = \dfrac{\sqrt{2}}{2}$ for $x=\dfrac{\pi}{4}$ and $x=-\dfrac{\pi}{4} = \dfrac{7\pi}{4}$; it is less than or equal to $\dfrac{\sqrt{2}}{2}$ to the left of the vertical line $x=\dfrac{\sqrt{2}}{2}$, that is on $\left[\dfrac{\pi}{4}, \dfrac{7\pi}{4}\right]$.

Sign table on $[0,2\pi)$:

$x$	$\left[0,\tfrac{\pi}{6}\right)$	$\left[\tfrac{\pi}{6},\tfrac{\pi}{4}\right)$	$\left[\tfrac{\pi}{4},\tfrac{5\pi}{6}\right]$	$\left(\tfrac{5\pi}{6},\tfrac{7\pi}{4}\right]$	$\left(\tfrac{7\pi}{4},2\pi\right)$
$\sin x - \tfrac{1}{2}$	$-$	$+$	$+$	$-$	$-$
$\tfrac{\sqrt{2}}{2}-\cos x$	$-$	$-$	$+$	$+$	$-$
product	$+$	$-$	$+$	$-$	$+$

The inequality $\ge 0$ is satisfied where the product is positive or zero: $\left[0,\frac{\pi}{6}\right] \cup \left[\frac{\pi}{4},\frac{5\pi}{6}\right] \cup \left[\frac{7\pi}{4},2\pi\right)$ Generalising with periodicity ($x \to x + 2k\pi$) gives the complete solution.

Links

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