Trigonometric inequalities

Inequalities with sin, cos, tan and cot, double inequalities, quadratic inequalities and systems. 12 problems with step-by-step solutions.

Solve each problem in your notebook first, then open the solution. In all answers \(k, m \in \mathbb{Z}\).

Problem 1 sin x and a number

Solve.

a) \(\sin x > \frac{\sqrt2}{2}\) b) \(\sin x \ge \frac12\) c) \(\sin x > -\frac12\) d) \(\sin x < \frac12\) e) \(\sin x \le -3\) f) \(\sin x > -5\)

Show solution

Look at the unit circle: \(\sin x > a\) is the arc above the horizontal line \(y = a\), from \(\arcsin a\) to \(\pi - \arcsin a\).

a) \(\frac{\pi}{4} + 2k\pi < x < \frac{3\pi}{4} + 2k\pi\)
b) \(\frac{\pi}{6} + 2k\pi \le x \le \frac{5\pi}{6} + 2k\pi\)
c) \(-\frac{\pi}{6} + 2k\pi < x < \frac{7\pi}{6} + 2k\pi\)
d) \(\frac{5\pi}{6} + 2k\pi < x < \frac{13\pi}{6} + 2k\pi\)
e) \(\sin x \ge -1 > -3\): no solution.
f) always true: \(x \in \mathbb{R}\).

Problem 2 sin of a multiple angle

Solve.

a) \(2\sin 3x \le \sqrt3\) b) \(4\sin 2x + 1 < 3\) c) \(2\sin 2x > \sqrt3\) d) \(\sin\left(3x - \frac{\pi}{4}\right) \le 0\) e) \(2\sin(-x) > 1\) f) \(2\sin x \ge -\sqrt2\)

Show solution

a) \(\sin 3x \le \frac{\sqrt3}{2}\): \(\frac{2\pi}{3} + 2k\pi \le 3x \le \frac{7\pi}{3} + 2k\pi\), so \(\frac{2\pi}{9} + \frac{2k\pi}{3} \le x \le \frac{7\pi}{9} + \frac{2k\pi}{3}\).

b) \(\sin 2x < \frac12\): \(\frac{5\pi}{6} + 2k\pi < 2x < \frac{13\pi}{6} + 2k\pi\), so \(\frac{5\pi}{12} + k\pi < x < \frac{13\pi}{12} + k\pi\).

c) \(\sin 2x > \frac{\sqrt3}{2}\): \(\frac{\pi}{3} + 2k\pi < 2x < \frac{2\pi}{3} + 2k\pi\), so \(\frac{\pi}{6} + k\pi < x < \frac{\pi}{3} + k\pi\).

d) \(-\pi + 2k\pi \le 3x - \frac{\pi}{4} \le 2k\pi\), so \(-\frac{\pi}{4} + \frac{2k\pi}{3} \le x \le \frac{\pi}{12} + \frac{2k\pi}{3}\).

e) \(\sin x < -\frac12\): \(\frac{7\pi}{6} + 2k\pi < x < \frac{11\pi}{6} + 2k\pi\).

f) \(\sin x \ge -\frac{\sqrt2}{2}\): \(-\frac{\pi}{4} + 2k\pi \le x \le \frac{5\pi}{4} + 2k\pi\).

Problem 3 cos x and a number

Solve.

a) \(\cos x \ge \frac12\) b) \(\cos x < 0\) c) \(2\cos 2x + 1 > 0\) d) \(2\cos x \ge \sqrt3\) e) \(3\cos 2x > -\pi\)

Show solution

\(\cos x > a\) is the arc to the right of the vertical line \(x = a\): \(-\arccos a < x < \arccos a\) (plus \(2k\pi\)).

a) \(-\frac{\pi}{3} + 2k\pi \le x \le \frac{\pi}{3} + 2k\pi\)
b) \(\frac{\pi}{2} + 2k\pi < x < \frac{3\pi}{2} + 2k\pi\)
c) \(\cos 2x > -\frac12\): \(-\frac{2\pi}{3} + 2k\pi < 2x < \frac{2\pi}{3} + 2k\pi\), so \(-\frac{\pi}{3} + k\pi < x < \frac{\pi}{3} + k\pi\)
d) \(\cos x \ge \frac{\sqrt3}{2}\): \(-\frac{\pi}{6} + 2k\pi \le x \le \frac{\pi}{6} + 2k\pi\)
e) \(\cos 2x > -\frac{\pi}{3} \approx -1.05\), which is always true: \(x \in \mathbb{R}\)

Problem 4 cos of a multiple angle

Solve.

a) \(2\cos\left(2x + \frac{\pi}{3}\right) \le 1\) b) \(\cos 2x > \frac{\sqrt2}{2}\) c) \(\cos(-3x) < \frac12\) d) \(3\cos 2x - 1 > 0\)

Show solution

a) \(\cos\left(2x + \frac{\pi}{3}\right) \le \frac12\): \(\frac{\pi}{3} + 2k\pi \le 2x + \frac{\pi}{3} \le \frac{5\pi}{3} + 2k\pi\), so \(k\pi \le x \le \frac{2\pi}{3} + k\pi\).

b) \(-\frac{\pi}{4} + 2k\pi < 2x < \frac{\pi}{4} + 2k\pi\), so \(-\frac{\pi}{8} + k\pi < x < \frac{\pi}{8} + k\pi\).

c) \(\cos 3x < \frac12\): \(\frac{\pi}{3} + 2k\pi < 3x < \frac{5\pi}{3} + 2k\pi\), so \(\frac{\pi}{9} + \frac{2k\pi}{3} < x < \frac{5\pi}{9} + \frac{2k\pi}{3}\).

d) \(\cos 2x > \frac13\): \(-\arccos\frac13 + 2k\pi < 2x < \arccos\frac13 + 2k\pi\), so \(-\frac12\arccos\frac13 + k\pi < x < \frac12\arccos\frac13 + k\pi\).

Problem 5 Inequalities in [0, 2π]

Solve in the interval \([0, 2\pi]\).

a) \(2\cos x - \sqrt2 \ge 0\) b) \(\sin\left(2x - \frac{\pi}{6}\right) \ge \frac12\) c) \(2\sin x + 1 > 0\) d) \(\cos x - \sin x > 0\)

Show solution

a) \(\cos x \ge \frac{\sqrt2}{2}\): \(x \in \left[0, \frac{\pi}{4}\right] \cup \left[\frac{7\pi}{4}, 2\pi\right]\).

b) \(\frac{\pi}{6} + 2k\pi \le 2x - \frac{\pi}{6} \le \frac{5\pi}{6} + 2k\pi\), so \(\frac{\pi}{6} + k\pi \le x \le \frac{\pi}{2} + k\pi\). In \([0, 2\pi]\): \(\left[\frac{\pi}{6}, \frac{\pi}{2}\right] \cup \left[\frac{7\pi}{6}, \frac{3\pi}{2}\right]\).

c) \(\sin x > -\frac12\): \(x \in \left[0, \frac{7\pi}{6}\right) \cup \left(\frac{11\pi}{6}, 2\pi\right]\).

d) \(\sqrt2\cos\left(x + \frac{\pi}{4}\right) > 0\): \(-\frac{\pi}{2} + 2k\pi < x + \frac{\pi}{4} < \frac{\pi}{2} + 2k\pi\), so \(-\frac{3\pi}{4} + 2k\pi < x < \frac{\pi}{4} + 2k\pi\). In \([0, 2\pi]\): \(\left[0, \frac{\pi}{4}\right) \cup \left(\frac{5\pi}{4}, 2\pi\right]\).

Problem 6 tan x and a number

Solve.

a) \(\tan x > \sqrt3\) b) \(\tan x \le 1\) c) \(3\tan\left(2x - \frac{\pi}{6}\right) - \sqrt3 \ge 0\) d) \(\tan 2x < 1\) e) \(2\tan 3x < 2\sqrt3\) f) \(\tan\left(x + \frac{\pi}{4}\right) \ge 1\)

Show solution

The tangent is increasing on \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) and has period \(\pi\): \(\tan x > a \iff \arctan a + k\pi < x < \frac{\pi}{2} + k\pi\).

a) \(\frac{\pi}{3} + k\pi < x < \frac{\pi}{2} + k\pi\)
b) \(-\frac{\pi}{2} + k\pi < x \le \frac{\pi}{4} + k\pi\)
c) \(\tan\left(2x - \frac{\pi}{6}\right) \ge \frac{\sqrt3}{3}\): \(\frac{\pi}{6} + k\pi \le 2x - \frac{\pi}{6} < \frac{\pi}{2} + k\pi\), so \(\frac{\pi}{6} + \frac{k\pi}{2} \le x < \frac{\pi}{3} + \frac{k\pi}{2}\)
d) \(-\frac{\pi}{2} + k\pi < 2x < \frac{\pi}{4} + k\pi\), so \(-\frac{\pi}{4} + \frac{k\pi}{2} < x < \frac{\pi}{8} + \frac{k\pi}{2}\)
e) \(\tan 3x < \sqrt3\): \(-\frac{\pi}{2} + k\pi < 3x < \frac{\pi}{3} + k\pi\), so \(-\frac{\pi}{6} + \frac{k\pi}{3} < x < \frac{\pi}{9} + \frac{k\pi}{3}\)
f) \(\frac{\pi}{4} + k\pi \le x + \frac{\pi}{4} < \frac{\pi}{2} + k\pi\), so \(k\pi \le x < \frac{\pi}{4} + k\pi\)

Problem 7 cot x and a number

Solve.

a) \(\cot x > \sqrt3\) b) \(3\cot 2x - \sqrt3 \le 0\) c) \(\cot 3x \ge 1\) d) \(2\cot 2x - 2 < 0\) e) \(\sqrt3\cot\left(x - \frac{\pi}{6}\right) \ge 1\) f) \(\tan x \ge 1\) in \([0, 2\pi]\)

Show solution

The cotangent is decreasing on \((0, \pi)\) with period \(\pi\): \(\cot x > a \iff k\pi < x < \operatorname{arccot} a + k\pi\).

a) \(k\pi < x < \frac{\pi}{6} + k\pi\)
b) \(\cot 2x \le \frac{\sqrt3}{3}\): \(\frac{\pi}{3} + k\pi \le 2x < \pi + k\pi\), so \(\frac{\pi}{6} + \frac{k\pi}{2} \le x < \frac{\pi}{2} + \frac{k\pi}{2}\)
c) \(k\pi < 3x \le \frac{\pi}{4} + k\pi\), so \(\frac{k\pi}{3} < x \le \frac{\pi}{12} + \frac{k\pi}{3}\)
d) \(\cot 2x < 1\): \(\frac{\pi}{4} + k\pi < 2x < \pi + k\pi\), so \(\frac{\pi}{8} + \frac{k\pi}{2} < x < \frac{\pi}{2} + \frac{k\pi}{2}\)
e) \(\cot\left(x - \frac{\pi}{6}\right) \ge \frac{\sqrt3}{3}\): \(k\pi < x - \frac{\pi}{6} \le \frac{\pi}{3} + k\pi\), so \(\frac{\pi}{6} + k\pi < x \le \frac{\pi}{2} + k\pi\)
f) \(x \in \left[\frac{\pi}{4}, \frac{\pi}{2}\right) \cup \left[\frac{5\pi}{4}, \frac{3\pi}{2}\right)\)

Problem 8 cot in an interval

Find all solutions of \(\cot 2x \ge -1\) in \([0, 2\pi]\).

Show solution

\(k\pi < 2x \le \frac{3\pi}{4} + k\pi\), so \(\frac{k\pi}{2} < x \le \frac{3\pi}{8} + \frac{k\pi}{2}\).

In \([0, 2\pi]\) (\(k = 0, 1, 2, 3\)): \(\left(0, \frac{3\pi}{8}\right] \cup \left(\frac{\pi}{2}, \frac{7\pi}{8}\right] \cup \left(\pi, \frac{11\pi}{8}\right] \cup \left(\frac{3\pi}{2}, \frac{15\pi}{8}\right]\).

Problem 9 Double inequalities

Solve.

a) \(-\frac{\sqrt2}{2} < \cos x \le \frac12\) b) \(\frac12 < \sin x < \frac{\sqrt2}{2}\) c) \(-\frac12 < \sin x \le \frac{\sqrt3}{2}\)

d) \(|\tan x| < 1\) e) \(|\tan x| \ge \sqrt3\) f) \(1 \le \tan x \le \sqrt3\) g) \(\tan\frac{x}{2} > -1\)

Show solution

a) On \([0, 2\pi)\): \(\cos x \le \frac12\) for \(x \in \left[\frac{\pi}{3}, \frac{5\pi}{3}\right]\), and \(\cos x > -\frac{\sqrt2}{2}\) excludes \(\left[\frac{3\pi}{4}, \frac{5\pi}{4}\right]\). Answer: \(\left[\frac{\pi}{3} + 2k\pi, \frac{3\pi}{4} + 2k\pi\right) \cup \left(\frac{5\pi}{4} + 2k\pi, \frac{5\pi}{3} + 2k\pi\right]\).

b) \(\left(\frac{\pi}{6} + 2k\pi, \frac{\pi}{4} + 2k\pi\right) \cup \left(\frac{3\pi}{4} + 2k\pi, \frac{5\pi}{6} + 2k\pi\right)\).

c) \(\sin x > -\frac12\) on \(\left(-\frac{\pi}{6}, \frac{7\pi}{6}\right)\); remove \(\left(\frac{\pi}{3}, \frac{2\pi}{3}\right)\) where \(\sin x > \frac{\sqrt3}{2}\). Answer: \(\left(-\frac{\pi}{6} + 2k\pi, \frac{\pi}{3} + 2k\pi\right] \cup \left[\frac{2\pi}{3} + 2k\pi, \frac{7\pi}{6} + 2k\pi\right)\).

d) \(-1 < \tan x < 1\): \(-\frac{\pi}{4} + k\pi < x < \frac{\pi}{4} + k\pi\).

e) \(\frac{\pi}{3} + k\pi \le x \le \frac{2\pi}{3} + k\pi\), \(x \ne \frac{\pi}{2} + k\pi\).

f) \(\frac{\pi}{4} + k\pi \le x \le \frac{\pi}{3} + k\pi\).

g) \(-\frac{\pi}{4} + k\pi < \frac{x}{2} < \frac{\pi}{2} + k\pi\), so \(-\frac{\pi}{2} + 2k\pi < x < \pi + 2k\pi\).

Problem 10 Quadratic inequalities

Solve.

a) \(2\sin^2 x - \sin x - 1 \le 0\) b) \(4\cos^2 x - 1 < 0\) c) \(2\sin^2 x - 3\sin x + 1 > 0\) d) \(\tan^2 x - \tan x - 2 < 0\) e) \(\sin^2 x > \frac34\) f) \(\cos 2x + 3\cos x \ge 1\)

Show solution

a) \((2\sin x + 1)(\sin x - 1) \le 0 \iff -\frac12 \le \sin x \le 1\), so \(-\frac{\pi}{6} + 2k\pi \le x \le \frac{7\pi}{6} + 2k\pi\).

b) \(|\cos x| < \frac12\): \(\frac{\pi}{3} + k\pi < x < \frac{2\pi}{3} + k\pi\).

c) \((2\sin x - 1)(\sin x - 1) > 0 \iff \sin x < \frac12\) (because \(\sin x > 1\) is impossible): \(\frac{5\pi}{6} + 2k\pi < x < \frac{13\pi}{6} + 2k\pi\).

d) \((\tan x - 2)(\tan x + 1) < 0 \iff -1 < \tan x < 2\): \(-\frac{\pi}{4} + k\pi < x < \arctan 2 + k\pi\).

e) \(|\sin x| > \frac{\sqrt3}{2}\): \(\frac{\pi}{3} + k\pi < x < \frac{2\pi}{3} + k\pi\).

f) \(2\cos^2 x + 3\cos x - 2 \ge 0 \iff (2\cos x - 1)(\cos x + 2) \ge 0 \iff \cos x \ge \frac12\): \(-\frac{\pi}{3} + 2k\pi \le x \le \frac{\pi}{3} + 2k\pi\).

Problem 11 Using formulas

Solve.

a) \((\sin x + \cos x)^2 < \frac32\) b) \(\sin 3x\cos x - \cos 3x\sin x > \frac12\) c) \(\cos^2 x - \sin^2 x > \frac12\)

Show solution

a) \(1 + \sin 2x < \frac32\), so \(\sin 2x < \frac12\): \(\frac{5\pi}{12} + k\pi < x < \frac{13\pi}{12} + k\pi\).

b) \(\sin 2x > \frac12\): \(\frac{\pi}{6} + 2k\pi < 2x < \frac{5\pi}{6} + 2k\pi\), so \(\frac{\pi}{12} + k\pi < x < \frac{5\pi}{12} + k\pi\).

c) \(\cos 2x > \frac12\): \(-\frac{\pi}{6} + k\pi < x < \frac{\pi}{6} + k\pi\).

Problem 12 Systems of inequalities

Solve.

a) \(\begin{cases}\sin x < 0\\ \cos x > 0\end{cases}\) b) \(\begin{cases}\sin x \ge \frac12\\ \cos x \le 0\end{cases}\) c) \(\begin{cases}\tan x > -1\\ \cot x > \frac{\sqrt3}{3}\end{cases}\) d) \(\begin{cases}\sin x \le \frac12\\ \cos x > \frac{\sqrt2}{2}\end{cases}\)

Show solution

a) Fourth quadrant: \(\frac{3\pi}{2} + 2k\pi < x < 2\pi + 2k\pi\).

b) \(\sin x \ge \frac12\) on \(\left[\frac{\pi}{6}, \frac{5\pi}{6}\right]\), \(\cos x \le 0\) on \(\left[\frac{\pi}{2}, \frac{3\pi}{2}\right]\): \(\frac{\pi}{2} + 2k\pi \le x \le \frac{5\pi}{6} + 2k\pi\).

c) \(\cot x > \frac{\sqrt3}{3}\) on \(\left(k\pi, \frac{\pi}{3} + k\pi\right)\); there \(\tan x > \sqrt3 > -1\), so the first inequality holds too. Answer: \(k\pi < x < \frac{\pi}{3} + k\pi\).

d) \(\cos x > \frac{\sqrt2}{2}\) on \(\left(-\frac{\pi}{4}, \frac{\pi}{4}\right)\); there \(\sin x \le \frac12\) means \(x \le \frac{\pi}{6}\). Answer: \(-\frac{\pi}{4} + 2k\pi < x \le \frac{\pi}{6} + 2k\pi\).