Inequality reducible to the cosine

Problem

Solve the inequality: $2\cos^2 x - 1 < 0$

Solution

First route (double-angle formula). We recognise $2\cos^2 x - 1 = \cos(2x)$, so the inequality becomes $\cos(2x) < 0$. The cosine is negative on the arcs $\left(\dfrac{\pi}{2}+2k\pi,\ \dfrac{3\pi}{2}+2k\pi\right)$; putting $2x$ in place of the argument: $\frac{\pi}{2}+2k\pi < 2x < \frac{3\pi}{2}+2k\pi \iff \frac{\pi}{4}+k\pi < x < \frac{3\pi}{4}+k\pi$

Second route (direct study). $2\cos^2 x - 1 < 0 \iff \cos^2 x < \dfrac{1}{2} \iff |\cos x| < \dfrac{\sqrt{2}}{2}$. On the circle $\cos x = \pm\dfrac{\sqrt{2}}{2}$ at $\dfrac{\pi}{4}, \dfrac{3\pi}{4}, \dfrac{5\pi}{4}, \dfrac{7\pi}{4}$; the condition is satisfied on $\left(\dfrac{\pi}{4},\dfrac{3\pi}{4}\right)\cup\left(\dfrac{5\pi}{4},\dfrac{7\pi}{4}\right)$, in agreement with the first route. $\boxed{\;x\in\left(\frac{\pi}{4}+k\pi,\ \frac{3\pi}{4}+k\pi\right)\;}$

Links

Topics: Trigonometric inequalities
Concepts: Trigonometric inequality · Double-angle formulae · Absolute value
Functions: Cosine
Skills: Solving inequalities · Using formulae
Exercise type: Solving an inequality