Topic 3.10A

Trigonometric Equations and Inequalities

In 3.9 an inverse trig function gave us one angle. But a trig equation usually has more solutions than that: two in every cycle, and infinitely many overall. This note (Part A) shows how to find all of them. Part B uses the same ideas to solve inequalities.

9 MIN READ5 IDEAS33 PROBLEMS10 flashcards

Practice 33 problems Next: Chapter 3.10B Trigonometric Equations and Inequalities

Two Solutions per Cycle

Every trig equation is solved in the same order: first isolate the trig function, then find the angles.

Worked example

Example 1. Solve $3 - 2 \sin \theta = 4$ for $0 \le \theta < 2\pi$ . Then give all solutions.

01
Isolate $\sin \theta$ , as you would isolate $x$ : $-2 \sin \theta = 1$ , so $\sin \theta = -1/2$ .
02
Find the reference angle: $\sin (\pi /6) = 1/2$ , so the reference angle is $\pi /6$ .
03
Sine is the y-coordinate, and it is negative in Quadrants III and IV. The angles with reference angle $\pi /6$ there are $\pi + \pi /6 = 7\pi /6$ and $2\pi - \pi /6 = 11\pi /6$ .
04
Both lie in $[0, 2\pi )$ , so the solutions are $\theta = 7\pi /6$ and $\theta = 11\pi /6$ .

The horizontal line $y = -1/2$ meets the unit circle, and one cycle of the sine graph, exactly twice.

Because sine repeats every $2\pi$ , adding any whole number of full turns gives another solution. The complete solution set is:
$\theta=\frac{7\pi}{6}+2\pi k\quad\mathrm{or}\quad\theta=\frac{11\pi}{6}+2\pi k$
where $k$ is any integer. Graphically, the line $y = -1/2$ crosses the sine graph twice in every cycle, and each crossing repeats every $2\pi$ .

COMMON MISTAKE

" $\sin \theta = -1/2$ , so $\theta = \arcsin (-1/2) = -\pi /6$ . Done."

$\arcsin$ returns only ONE angle, the one in $[-\pi /2, \pi /2]$ (3.9). $-\pi /6$ is a correct solution, but it is not in $[0, 2\pi )$ , and it is only one of the two solutions per cycle.

From $-\pi /6$ you can reach both answers: add $2\pi$ to get $11\pi /6$ , and use symmetry $(\pi - (-\pi /6))$ to get $7\pi /6$ . Either way, always look for the second angle.

Quick check

Solve $2\cos\theta = 1$ for $0 \le \theta < 2\pi$ .

Using Inverse Trig Functions

When the value is not a special one, use a calculator in RADIAN mode. The inverse function gives the first angle; symmetry gives the second.

CONCEPT

Finding both angles in one cycle

$\sin \theta = k$ : $\theta = \arcsin k$ or $\theta = \pi - \arcsin k$ (the two angles are symmetric about $\pi /2$ ).

$\cos \theta = k$ : $\theta = \arccos k$ or $\theta = -\arccos k$ (symmetric about the x-axis); add $2\pi$ to bring $-\arccos k$ into $[0, 2\pi )$ .

$\tan \theta = k$ : $\theta = \arctan k$ , and then every $\pi$ after it (tangent has period $\pi$ , 3.8).

Worked example

Example 2. Solve $5 \cos \theta - 1 = 2$ for $0 \le \theta < 2\pi$ . Round to three decimal places.

01
Isolate: $5 \cos \theta = 3$ , so $\cos \theta = 3/5 = 0.6$ .
02
First angle: $\theta = \arccos (0.6) \approx 0.927$ .
03
Cosine is the x-coordinate. The other point on the unit circle with $x = 0.6$ is the mirror image below the x-axis, at −0.927. In $[0, 2\pi )$ that is $2\pi - 0.927 \approx 5.356$ .

The vertical line $x = 0.6$ meets the unit circle at two angles, symmetric about the x-axis.

COMMON MISTAKE

"The second solution of $\cos \theta = 0.6$ is $\pi - 0.927 \approx 2.214$ ."

$\pi - \theta$ is the partner angle for SINE, because it keeps the y-coordinate. It flips the x-coordinate: $\cos (2.214) \approx -0.6$ , not 0.6.

Cosine keeps the x-coordinate, so its partner is the reflection over the x-axis: $-\theta$ , or $2\pi - \theta$ .

Worked example

Example 3. Solve $2 \tan \theta + 5 = 0$ for $0 \le \theta < 2\pi$ .

01
Isolate: $\tan \theta = -2.5$ . Then $\arctan (-2.5) \approx -1.190$ , which is not in $[0, 2\pi )$ .
02
Tangent repeats every $\pi$ , so add $\pi$ until the angles land in the interval: $-1.190 + \pi \approx 1.951$ and $-1.190 + 2\pi \approx 5.093$ .
03
Check: $\tan \theta$ is negative in Quadrants II and IV, and 1.951 (between $\pi /2$ and $\pi$ ) and 5.093 (between $3\pi /2$ and $2\pi$ ) are in exactly those quadrants ✓.

Equations That Factor

Sometimes the trig function appears squared, or more than once. Treat the trig function like a single variable and factor, just like a quadratic from Unit 1.

Worked example

Example 4. Solve $2 \cos ^{2} \theta + \cos \theta - 1 = 0$ for $0 \le \theta < 2\pi$ .

01
Think of $\cos \theta$ as $c$ : $2c^{2} + c - 1 = (2c - 1)(c + 1)$ . So $(2 \cos \theta - 1)(\cos \theta + 1) = 0$ .
02
Zero product property: $\cos \theta = 1/2$ or $\cos \theta = -1$ .
03
$\cos \theta = 1/2$ gives $\theta = \pi /3$ and $5\pi /3$ . $\cos \theta = -1$ gives only $\theta = \pi$ (the one point $(-1, 0)$ ).
04
Solutions: $\theta = \pi /3$ , $\pi$ , and $5\pi /3$ .

COMMON MISTAKE

Solving $3 \sin \theta \cos \theta = \sin \theta$ by dividing both sides by $\sin \theta$ to get $\cos \theta = 1/3$ .

Dividing by $\sin \theta$ is only allowed when $\sin \theta \ne 0$ , so this throws away every solution with $\sin \theta = 0$ .

Instead, move everything to one side and factor: $\sin \theta$ $(3 \cos \theta - 1) = 0$ . Now $\sin \theta = 0$ gives $\theta = 0$ and $\pi$ , and $\cos \theta = 1/3$ gives $\theta \approx 1.231$ and 5.052. That is four solutions, not two.

When the Period Changes

If the input is $b\theta$ instead of $\theta$ , the graph completes more cycles in $[0, 2\pi )$ , so there are more solutions. Remember from 3.6A: $y = \sin 2\theta$ has period $\pi$ , so it fits two full cycles into $[0, 2\pi )$ .

Worked example

Example 5. Solve $2 \sin$ $2\theta = \sqrt{3}$ for $0 \le \theta < 2\pi$ .

01
Isolate: $\sin$ $2\theta = \sqrt{3}/2$ . Let $u = 2\theta$ . Because $0 \le \theta < 2\pi$ , the new variable runs over $0 \le u < 4\pi$ : two full turns.
02
In the first turn, $\sin u = \sqrt{3}/2$ at $u = \pi /3$ and $2\pi /3$ . Add $2\pi$ for the second turn: $u = 7\pi /3$ and $8\pi /3$ .
03
Divide every $u$ by 2: $\theta = \pi /6$ , $\pi /3$ , $7\pi /6$ , and $4\pi /3$ .

$\sin \theta$ reaches $\sqrt{3}/2$ twice in $[0, 2\pi )$ , but $\sin 2\theta$ does it four times.

COMMON MISTAKE

" $\sin$ $2\theta = \sqrt{3}/2$ , so $2\theta = \pi /3$ or $2\pi /3$ , and $\theta = \pi /6$ or $\pi /3$ ."

That solves only the first of the two cycles. The interval for $u = 2\theta$ is twice as long as the interval for $\theta$ , so you must collect u-values from $[0, 4\pi )$ BEFORE dividing by 2. Check with the graph: the curve $y = \sin 2\theta$ clearly meets the line four times.

REAL-LIFE EXAMPLE

Every time the pedal is at 40 cm

Remember from 3.9: $h(t) = -17 \cos (5\pi t/3) + 28 = 40$ led to $\cos (5\pi t/3) = -12/17$ , and $\arccos$ gave the first time, $t \approx 0.45$ s. Now we can find them all.

Let $u = 5\pi t/3$ . $\cos u = -12/17$ gives $u = \arccos (-12/17) \approx 2.354$ and $u = 2\pi - 2.354 \approx 3.929$ . Then $t = 3u/(5\pi )$ : $t \approx 0.45$ s (on the way up) and $t \approx 0.75$ s (on the way down).

One rotation takes 1.2 s, so all the times are $t \approx 0.45 + 1.2k$ and $t \approx 0.75 + 1.2k$ for $k = 0$ , 1, 2, …

Check with symmetry: the top of the rotation is at $t = 0.6$ , and 0.45 and 0.75 are both 0.15 s away from it ✓.

The pedal passes 40 cm twice per rotation: once rising, once falling.

Quick check

How many solutions does $\sin(2\theta) = 0$ have on $0 \le \theta < 2\pi$ ?

Common slips

" $\sin \theta = -1/2$ , so $\theta = \arcsin (-1/2) = -\pi /6$ . Done."

$\arcsin$ returns only one angle, the one in $[-\pi /2, \pi /2]$ (3.9). $-\pi /6$ is a correct solution, but it is not in $[0, 2\pi )$ , and it is only one of the two solutions per cycle.

From $-\pi /6$ you can reach both answers: add $2\pi$ to get $11\pi /6$ , and use symmetry $(\pi - (-\pi /6))$ to get $7\pi /6$ . Either way, always look for the second angle.
"The second solution of $\cos \theta = 0.6$ is $\pi - 0.927 \approx 2.214$ ."

$\pi - \theta$ is the partner angle for sine, because it keeps the y-coordinate. It flips the x-coordinate: $\cos (2.214) \approx -0.6$ , not 0.6.

Cosine keeps the x-coordinate, so its partner is the reflection over the x-axis: $-\theta$ , or $2\pi - \theta$ .
Solving $3 \sin \theta \cos \theta = \sin \theta$ by dividing both sides by $\sin \theta$ to get $\cos \theta = 1/3$ .

Dividing by $\sin \theta$ is only allowed when $\sin \theta \ne 0$ , so this throws away every solution with $\sin \theta = 0$ .

Instead, move everything to one side and factor: $\sin \theta$ $(3 \cos \theta - 1) = 0$ . Now $\sin \theta = 0$ gives $\theta = 0$ and $\pi$ , and $\cos \theta = 1/3$ gives $\theta \approx 1.231$ and 5.052. That is four solutions, not two.
" $\sin$ $2\theta = \sqrt{3}/2$ , so $2\theta = \pi /3$ or $2\pi /3$ , and $\theta = \pi /6$ or $\pi /3$ ."

That solves only the first of the two cycles. The interval for $u = 2\theta$ is twice as long as the interval for $\theta$ , so you must collect u-values from $[0, 4\pi )$ Before dividing by 2. Check with the graph: the curve $y = \sin 2\theta$ clearly meets the line four times.

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10 cards · Inverse trig and equations

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Recap card

6 lines to re-read the night before.

01
Isolate the trig function first, then find the angles.
02
Most equations have two solutions per cycle. For sine the partner of $\theta$ is $\pi - \theta$ ; for cosine it is $-\theta$ (or $2\pi - \theta$ ). Tangent repeats every $\pi$ .
03
An inverse trig function gives only one angle. Use symmetry and periodicity to find the rest.
04
All solutions: add $2\pi k$ (or $\pi k$ for tangent), where $k$ is any integer.
05
Factor instead of dividing by a trig function, so no solutions are lost.
06
With input $b\theta$ , collect solutions over the stretched interval first, then divide by $b$ .

Practice 33 problems Next: Chapter 3.10B Trigonometric Equations and Inequalities

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Trigonometric Equations and Inequalities

Two Solutions per Cycle

Using Inverse Trig Functions

Equations That Factor

When the Period Changes

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