Absolute value of the cosine

Problem

Solve the inequality: $|\cos x| > \frac{\sqrt{2}}{2}$

Solution

It is an inequality $|A| > k$ with $k>0$, which is equivalent to $\cos x > \dfrac{\sqrt{2}}{2}$ or $\cos x < -\dfrac{\sqrt{2}}{2}$.

• $\cos x > \dfrac{\sqrt{2}}{2}$ (to the right of the line $x=\dfrac{\sqrt{2}}{2}$): on $\left[0,\dfrac{\pi}{4}\right)\cup\left(\dfrac{7\pi}{4},2\pi\right)$.
• $\cos x < -\dfrac{\sqrt{2}}{2}$ (to the left of the line $x=-\dfrac{\sqrt{2}}{2}$): on $\left(\dfrac{3\pi}{4},\dfrac{5\pi}{4}\right)$.

Combining the two cases, on $[0,2\pi)$: $\boxed{\;x\in\left[0,\frac{\pi}{4}\right)\cup\left(\frac{3\pi}{4},\frac{5\pi}{4}\right)\cup\left(\frac{7\pi}{4},2\pi\right)\;}$

Links

Topics: Trigonometric inequalities
Concepts: Trigonometric inequality · Absolute value
Functions: Cosine
Skills: Reasoning by cases · Solving inequalities
Exercise type: Solving an inequality