Topic 3.10B

Trigonometric Equations and Inequalities

An inequality asks for more than a few angles. It asks for every angle where one side is bigger than the other, and the answer is a set of intervals. The good news: the endpoints of those intervals are exactly the solutions of an equation, which you learned to find in 3.10A.

7 MIN READ5 IDEAS32 PROBLEMS10 flashcards

Read this first

30 sec

01
Solve the equation to find the boundaries. Then decide which pieces between them work.

Practice 32 problems Next: Chapter 3.11 The Secant, Cosecant, and Cotangent Functions

Boundary Points, Then Test

KEY RULE

Solve the equation to find the boundaries. Then decide which pieces between them work.

CONCEPT

Three steps for any trig inequality

Replace the inequality sign with = and solve the equation on the interval. These are the boundary points.
The boundary points cut the interval into pieces. On each piece the inequality is either always true or always false, because the graph can only change sides at a crossing.
Test one point in each piece, or read the graph, and keep the pieces that work. Include a boundary point only if the sign includes equality (≤ or ≥).

Worked example

Example 1. Solve $3 - 2 \sin \theta \ge 4$ for $0 \le \theta < 2\pi$ .

01
Isolate $\sin \theta$ : $-2 \sin \theta \ge 1$ . Divide by −2 and FLIP the sign: $\sin \theta \le -1/2$ .
02
Boundaries: from 3.10A Example 1, $\sin \theta = -1/2$ at $\theta = 7\pi /6$ and $11\pi /6$ . They cut $[0, 2\pi )$ into $[0, 7\pi /6)$ , $(7\pi /6, 11\pi /6)$ , and $(11\pi /6, 2\pi )$ .
03
Test a point in the middle piece: $\theta = 3\pi /2$ gives $3 - 2(-1) = 5 \ge 4$ ✓. Test $\theta = \pi /2$ in the first piece: $3 - 2(1) = 1$ , not ≥ 4 ✗. And $\theta = 23\pi /12$ in the last piece: $\sin (23\pi /12) \approx -0.26$ , which is not ≤ −1/2 ✗.
04
The sign is ≥, so the endpoints count. Solution: $7\pi /6 \le \theta \le 11\pi /6$ .

$y = 3 - 2 \sin \theta$ is on or above $y = 4$ exactly on $[7\pi /6, 11\pi /6]$ , shown on the θ-axis.

COMMON MISTAKE

Forgetting to flip the sign when dividing by −2, and solving $\sin \theta \ge -1/2$ instead.

Multiplying or dividing an inequality by a negative number reverses it, for trig functions just like for $x$ . The wrong inequality has exactly the opposite answer: every angle EXCEPT the interval $(7\pi /6, 11\pi /6)$ .

A single test point catches this: $\theta = \pi /2$ satisfies $\sin \theta \ge -1/2$ , but $3 - 2 \sin (\pi /2) = 1$ is not ≥ 4.

Quick check

Solve $\sin\theta > \frac{1}{2}$ on $[0, 2\pi)$ .

Asymptotes Are Boundaries Too

For tangent, the graph can also switch sides at a vertical asymptote without ever crossing the line. So the asymptotes must be added to the list of boundary points, and they are never included in the answer, because tangent is undefined there (3.8).

Worked example

Example 2. Solve $\tan \theta \ge 1$ for $0 \le \theta < 2\pi$ .

01
Equation: $\tan \theta = 1$ at $\theta = \pi /4$ and, one period later, at $5\pi /4$ .
02
Asymptotes in the interval: $\theta = \pi /2$ and $3\pi /2$ .
03
Tangent increases on each branch. So on the branch that starts at $\pi /2$ , it is below 1 until $5\pi /4$ and above 1 from $5\pi /4$ until the asymptote at $3\pi /2$ . The same happens on the first branch between $\pi /4$ and $\pi /2$ .
04
Solution: $\pi /4 \le \theta < \pi /2$ or $5\pi /4 \le \theta < 3\pi /2$ .

$\tan \theta \ge 1$ from each crossing up to the next asymptote. Open circles: the asymptotes are excluded.

COMMON MISTAKE

Writing the answer as $\pi /4 \le \theta \le \pi /2$ .

At $\theta = \pi /2$ the ray is vertical and $\tan \theta$ does not exist, so $\pi /2$ cannot satisfy any inequality. Asymptotes always get an open endpoint, even when the sign is ≤ or ≥.

Quick check

Solve $\tan\theta \ge 1$ on $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ .

When the Period Changes

Worked example

Example 3. Solve $\cos$ $2\theta > 1/2$ for $0 \le \theta < 2\pi$ .

01
Let $u = 2\theta$ , so $0 \le u < 4\pi$ (3.10A, Section 4). Equation: $\cos u = 1/2$ at $u = \pi /3$ , $5\pi /3$ , $7\pi /3$ , $11\pi /3$ .
02
$\cos u$ is above $1/2$ near its maxima, $u = 0$ , $2\pi$ , $4\pi$ . So $\cos u > 1/2$ for $0 \le u < \pi /3$ , $5\pi /3 < u < 7\pi /3$ , and $11\pi /3 < u < 4\pi$ .
03
Divide by 2: $0 \le \theta < \pi /6$ , $5\pi /6 < \theta < 7\pi /6$ , or $11\pi /6 < \theta < 2\pi$ . The sign is >, so the boundary points are excluded; $\theta = 0$ is included because it is a start of the interval, not a boundary.

$\cos$ $2\theta > 1/2$ on three intervals in $[0, 2\pi )$ , one around each maximum.

Inequalities in Context

Back in 3.7 we answered two "how long" questions with a calculator's intersect feature. Now they can be solved exactly.

REAL-LIFE EXAMPLE

The observation wheel

Remember from 3.7: $H(t) = -25 \cos (\pi t/10) + 28$ . When is the rider more than 40 m above the ground?

$H(t) > 40$ means $-25 \cos (\pi t/10) > 12$ , so $\cos (\pi t/10) < -12/25$ (the sign flips when dividing by −25).

Let $u = \pi t/10$ . In one turn, $\cos u = -12/25$ at $u = \arccos (-0.48) \approx 2.071$ and $u = 2\pi - 2.071$ . Cosine is below −0.48 between them, around its minimum at $u = \pi$ .

Convert with $t = 10u/\pi$ : $6.59 < t < 13.41$ . That is about 6.81 minutes per rotation, and it repeats every 20 minutes: $6.59 + 20k < t < 13.41 + 20k$ .

The same answer as the calculator in 3.7, now found algebraically.

REAL-LIFE EXAMPLE

The tide at the pier

Remember from 3.6B and 3.7: $D(t) = 1.8 \cos ((5\pi /31)(t - 3)) + 2.8$ . The water is too shallow for the boat when $D(t) < 2.0$ .

Isolate: $\cos ((5\pi /31)(t - 3)) < -4/9$ . One cycle of the input gives $\arccos (-4/9) \approx 2.031$ and $2\pi - 2.031$ , and cosine is below −4/9 between them.

Solve $(5\pi /31)(t - 3) = 2.031$ and = $2\pi - 2.031$ : $t \approx 7.01$ and $t \approx 11.39$ . The water is too shallow from about 7:01 a.m. to 11:23 a.m., matching 3.7.

Common slips

Forgetting to flip the sign when dividing by −2, and solving $\sin \theta \ge -1/2$ instead.

Multiplying or dividing an inequality by a negative number reverses it, for trig functions just like for $x$ . The wrong inequality has exactly the opposite answer: every angle except the interval $(7\pi /6, 11\pi /6)$ .

A single test point catches this: $\theta = \pi /2$ satisfies $\sin \theta \ge -1/2$ , but $3 - 2 \sin (\pi /2) = 1$ is not ≥ 4.
Writing the answer as $\pi /4 \le \theta \le \pi /2$ .

At $\theta = \pi /2$ the ray is vertical and $\tan \theta$ does not exist, so $\pi /2$ cannot satisfy any inequality. Asymptotes always get an open endpoint, even when the sign is ≤ or ≥.

Lock it in

Try the flashcards

10 cards · Inverse trig and equations

Start

Recap card

6 lines to re-read the night before.

01
Solve the related equation first: its solutions are the boundary points.
02
Between boundary points the inequality is always true or always false. Test one point per piece, or read the graph.
03
Flip the inequality sign when multiplying or dividing by a negative number.
04
For tangent, the asymptotes are extra boundary points and are always excluded.
05
With input $b\theta$ , find the intervals for $u = b\theta$ over the stretched interval, then divide by $b$ .
06
Include endpoints only for ≤ or ≥, and state context answers as time intervals.

Practice 32 problems Next: Chapter 3.11 The Secant, Cosecant, and Cotangent Functions

Unlock every question

Free

Premium

Sign in to FiveWay

Trigonometric Equations and Inequalities

Boundary Points, Then Test

Asymptotes Are Boundaries Too

When the Period Changes

Inequalities in Context

Recap card

Unlock Premium