Factorisation with the double-angle formula

Problem

Solve the inequality (hint: double-angle formula): $\sin(2x) + \cos x > 0$

Solution

Applying the double-angle formula $\sin(2x) = 2\sin x\cos x$ and factoring out $\cos x$: $2\sin x\cos x + \cos x = \cos x\,(2\sin x + 1) > 0$ We study the sign of the two factors on $[0,2\pi)$.

• $\cos x > 0$ on $\left[0,\dfrac{\pi}{2}\right)\cup\left(\dfrac{3\pi}{2},2\pi\right)$; it vanishes at $\dfrac{\pi}{2}, \dfrac{3\pi}{2}$.
• $2\sin x + 1 > 0 \iff \sin x > -\dfrac{1}{2}$, that is on $\left[0,\dfrac{7\pi}{6}\right)\cup\left(\dfrac{11\pi}{6},2\pi\right)$; it vanishes at $\dfrac{7\pi}{6}, \dfrac{11\pi}{6}$.

Sign table:

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

The inequality $>0$ is satisfied (excluding the zeros) on: $\boxed{\;x\in\left[0,\frac{\pi}{2}\right)\cup\left(\frac{7\pi}{6},\frac{3\pi}{2}\right)\cup\left(\frac{11\pi}{6},2\pi\right)\;}$ plus the corresponding arcs obtained by periodicity ($+2k\pi$).

Links

Topics: Trigonometric inequalities
Concepts: Trigonometric inequality · Double-angle formulae · Factorisation · Sign table
Functions: Cosine · Sine
Skills: Reasoning by cases · Solving inequalities · Using formulae
Exercise type: Solving an inequality