Inequalities with absolute value and roots

When absolute values or square roots also appear in a trigonometric inequality, no new method is needed: the same techniques already seen in Year 3 are applied, with the sole difference that the signs of sine, cosine and tangent are studied on the circle.

Remark — Revision from Year 3

When absolute values or square roots also appear in a trigonometric inequality, the same methods seen in Chapter 18 (Irrational inequalities) of Year 3 are applied: splitting into cases for the absolute value, and handling the two types $\sqrt{A(x)}\le B(x)$ (a system of three conditions) and $\sqrt{A(x)}\ge B(x)$ (a union of two cases). The fact that $A,B$ contain trigonometric functions does not change the method: the signs of sine, cosine and tangent are studied on the circle.

Example — Trigonometric inequality with absolute value

Solve $|\sin x| \le \dfrac{1}{2}$ on $[0,2\pi)$.

It is the inequality $|A| \le k$ with $k>0$, which is equivalent to $-\dfrac{1}{2} \le \sin x \le \dfrac{1}{2}$: a system of two elementary inequalities.

• $\sin x \le \dfrac{1}{2}$ on $\left[0,\dfrac{\pi}{6}\right]\cup\left[\dfrac{5\pi}{6},2\pi\right)$;
• $\sin x \ge -\dfrac{1}{2}$ on $\left[0,\dfrac{7\pi}{6}\right]\cup\left[\dfrac{11\pi}{6},2\pi\right)$.

Intersecting the two conditions: $\left[0,\frac{\pi}{6}\right]\cup\left[\frac{5\pi}{6},\frac{7\pi}{6}\right]\cup\left[\frac{11\pi}{6},2\pi\right)$

Links

Topics: Trigonometric inequalities
Concepts: Trigonometric inequality · Irrational inequality · Sign study · Absolute value
Functions: Sine
Skills: Reasoning by cases · Solving inequalities