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Lesson 14 — Trigonometric equations and inequalities

Solving equations and inequalities involving sine, cosine, and tangent. General solutions and solutions on finite intervals.

Used in: 1st year HS (15 years old) · Math II Japanese (ch. 三角関数) · Trigonometry — US precalc

\sin x = a \iff x = \arcsin a + 2k\pi \ \text{or}\ x = \pi - \arcsin a + 2k\pi, \ k \in \mathbb{Z}

General structure of solutions to basic trigonometric equations. Periodicity means each equation has infinitely many solutions; restricting to an interval (for example $[0, 2\pi)$ ) selects a finite count.

Choose your door

Rigorous notation, full derivation, hypotheses

Structure of solutions

Definition· General solutions of basic trigonometric equations

Let $a \in [-1, 1]$ .

Sine: $\sin x = a \iff x = \arcsin a + 2k\pi \quad \text{or} \quad x = \pi - \arcsin a + 2k\pi,\quad k \in \mathbb{Z}$ where $\arcsin: [-1, 1] \to [-\pi/2, \pi/2]$ is the principal inverse (range in Q1 or Q4).
Cosine: $\cos x = a \iff x = \pm\arccos a + 2k\pi,\quad k \in \mathbb{Z}$ where $\arccos: [-1, 1] \to [0, \pi]$ .
Tangent: $\tan x = a \iff x = \arctan a + k\pi,\quad k \in \mathbb{Z}$ with $\arctan: \mathbb{R} \to (-\pi/2, \pi/2)$ . The period of tangent is $\pi$ , not $2\pi$ .

"We close this section with one final example of a trigonometric inequality. To solve such an inequality, we begin by replacing the inequality with the corresponding equation, then check the resulting intervals." — Stitz–Zeager, Precalculus §10.7

"Solving a trigonometric equation requires the same techniques as solving any equation: isolate the variable, use factoring, use the quadratic formula, use identities, and set each factor equal to zero." — OpenStax Algebra and Trigonometry 2e §9.5

Symmetries on the unit circle

Geometry of solutions: $\sin\alpha = \sin(\pi - \alpha)$ (vertical symmetry); $\cos\alpha = \cos(-\alpha)$ (horizontal symmetry).

Strategy of reduction

Definition· Types of reducible equations

Type	Example	Strategy
Polynomial in $\sin$	$2\sin^2 x - \sin x - 1 = 0$	$u = \sin x$ , quadratic formula
Polynomial in $\cos$	$\cos^2 x = 1/4$	$u = \cos x$ , square root
$\sin x \pm \cos x = c$	$\sin x + \cos x = 1$	$R\sin(x + \varphi)$ with $R = \sqrt{a^2+b^2}$
Double angle	$\sin 2x = \sin x$	$2\sin x\cos x = \sin x$ ; factor
Sum-to-product	$\sin x + \sin 3x = 0$	$2\sin 2x\cos x = 0$

Worked examples

Five examples with increasing difficulty — from the most direct (basic equation on an interval) to real-world modeling (finding when tide wave crests occur). Each example cites its source: the original problem always comes from an open book.

Example— 1· Basic sine equation on an interval (application)

Problem: Solve $\sin x = \tfrac{1}{2}$ on $[0, 2\pi)$ .

Strategy: The equation is of basic type. Use the unit circle to identify which angles of the first rotation have sine equal to $1/2$ — exactly two, symmetric about the vertical axis.

Solution:

Identify the reference angle. The value $1/2$ is tabulated: $\sin(\pi/6) = 1/2$ . This is the reference angle in the first quadrant.
Use sine symmetry. On the circle, sine has the same value at $\pi/6$ and at $\pi - \pi/6 = 5\pi/6$ (second quadrant). In quadrants 3 and 4 sine is negative.
List the solutions in $[0, 2\pi)$ . $x = \frac{\pi}{6} \quad \text{or} \quad x = \frac{5\pi}{6}$

Check: $\sin(\pi/6) = 0.5$ and $\sin(5\pi/6) = \sin(\pi - \pi/6) = \sin(\pi/6) = 0.5$ . Correct.

Source. OpenStax — Algebra and Trigonometry 2e §9.5 — CC-BY 4.0 license.

Example— 2· General solution in ℝ (application)

Problem: Find all real solutions to $\cos x = -\tfrac{\sqrt 2}{2}$ .

Strategy: Identify the basic solution in the interval $[0, \pi]$ via $\arccos$ , then apply the symmetry $\cos(-x) = \cos x$ and periodicity $2\pi$ to generate all solutions.

Solution:

Principal solution. Since $\arccos(-\sqrt 2/2) = 3\pi/4$ , this is the angle in the second quadrant.
Symmetry. By the principle $\cos(-x) = \cos x$ , the angle $-3\pi/4$ (or equivalent $5\pi/4$ ) also works.
Generalize. Adding $2k\pi$ (period of cosine): $x = \tfrac{3\pi}{4} + 2k\pi \quad \text{or} \quad x = -\tfrac{3\pi}{4} + 2k\pi, \quad k \in \mathbb{Z}.$

Check: For $k = 0$ : $\cos(3\pi/4) = -\sqrt 2/2$ and $\cos(-3\pi/4) = \cos(3\pi/4) = -\sqrt 2/2$ . Correct.

Source. Stitz–Zeager — Precalculus §10.7 — CC-BY-NC-SA license.

Example— 3· Quadratic equation in sin x (intermediate)

Problem: Solve $2\sin^2 x - \sin x - 1 = 0$ on $[0, 2\pi)$ .

Strategy: Substitute $u = \sin x$ and solve the quadratic $2u^2 - u - 1 = 0$ . Then return to the angles corresponding to each root.

Solution:

Substitution. Let $u = \sin x$ . The equation becomes $2u^2 - u - 1 = 0$ .
Quadratic formula. $u = \dfrac{1 \pm \sqrt{1 + 8}}{4} = \dfrac{1 \pm 3}{4}$ . So $u = 1$ or $u = -1/2$ .
Return to $x$ .
- $\sin x = 1 \Rightarrow x = \pi/2$ .
- $\sin x = -1/2 \Rightarrow x = 7\pi/6$ or $x = 11\pi/6$ (third and fourth quadrants).
Solution set on $[0, 2\pi)$ : $\{\pi/2,\ 7\pi/6,\ 11\pi/6\}$ .

Check: $2\sin^2(\pi/2) - \sin(\pi/2) - 1 = 2 - 1 - 1 = 0$ . $2\sin^2(7\pi/6) - \sin(7\pi/6) - 1 = 2(1/4) - (-1/2) - 1 = 0$ . Correct.

Source. OpenStax — Algebra and Trigonometry 2e §9.5 — CC-BY 4.0 license.

Example— 4· Trigonometric inequality (advanced)

Problem: Solve $\sin x > \tfrac{\sqrt 3}{2}$ on $[0, 2\pi)$ .

Strategy: Convert the inequality to an equation to find boundary points; then inspect, on the unit circle, which arc has sine above $\sqrt 3/2$ .

Solution:

Associated equation. $\sin x = \sqrt 3/2$ has solutions $x = \pi/3$ and $x = 2\pi/3$ on $[0, 2\pi)$ .
Graph analysis. The graph of $\sin x$ on $[0, 2\pi)$ is a curve that reaches its maximum at $\pi/2$ . Above $\sqrt 3/2$ , it lies only between $\pi/3$ and $2\pi/3$ .
Solution set. Since the inequality is strict, the endpoints do not enter: $x \in \left(\tfrac{\pi}{3},\ \tfrac{2\pi}{3}\right).$

Check: At $x = \pi/2$ (middle of the interval): $\sin(\pi/2) = 1 > \sqrt 3/2$ . At $x = 0$ (outside): $\sin(0) = 0 < \sqrt 3/2$ . Correct.

Source. Stitz–Zeager — Precalculus §10.7 — CC-BY-NC-SA license.

Example— 5· Tides as a sine wave (real-world modeling)

Problem: The height of the tide at a port is modeled by $h(t) = 3 \sin\!\bigl(\tfrac{\pi t}{6}\bigr)$ meters, with $t$ in hours from 6 AM. At which times in the interval $[0, 12]$ hours does the tide reach $h = 1.5$ m?

Strategy: Equating the model to the desired value reduces the problem to a basic sine equation. Restricting to $[0, 12]$ corresponds to one full cycle of the argument.

Solution:

Equation. $3 \sin(\pi t/6) = 1.5 \Leftrightarrow \sin(\pi t/6) = 1/2$ .
Solve $\sin u = 1/2$ with $u = \pi t/6$ . The basic solutions on $[0, 2\pi)$ are $u = \pi/6$ and $u = 5\pi/6$ .
Return to $t$ . $\pi t/6 = \pi/6 \Rightarrow t = 1$ . $\pi t/6 = 5\pi/6 \Rightarrow t = 5$ .
Check the interval. Both $t = 1$ and $t = 5$ are in $[0, 12]$ .

$t = 1\ \text{h} \quad \text{or} \quad t = 5\ \text{h}.$

Check: $h(1) = 3 \sin(\pi/6) = 3 \cdot 0.5 = 1.5$ . $h(5) = 3 \sin(5\pi/6) = 3 \cdot 0.5 = 1.5$ . Correct.

Source. Active Calculus §0.7 — Boelkins · CC-BY-NC-SA.

Exercise list

40 exercises · 10 with worked solution (25%)

Application 34Understanding 4Modeling 1Challenge 1

Ex. 14.1Application
Solve $\sin x = \dfrac{1}{2}$ on $[0, 2\pi)$ . (Ans: $x = \pi/6$ or $x = 5\pi/6$ .)
Select the correct option
$x = \dfrac{\pi}{6}$ or $x = \dfrac{5\pi}{6}$ $x = \dfrac{\pi}{6}$ or $x = \dfrac{7\pi}{6}$ $x = \dfrac{\pi}{3}$ or $x = \dfrac{2\pi}{3}$ $x = \dfrac{\pi}{6}$ only
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The reference angle is $\arcsin(1/2) = \pi/6$ . Sine is positive in Q1 and Q2, so the solutions on $[0,2\pi)$ are $x = \pi/6$ and $x = \pi - \pi/6 = 5\pi/6$ .
Show step-by-step (with the why)
1. Calculate the reference angle: $\arcsin(1/2) = \pi/6$ .
2. Identify quadrants: sine positive in Q1 and Q2.
3. Q1: $x_1 = \pi/6$ .
4. Q2: $x_2 = \pi - \pi/6 = 5\pi/6$ .
Ex. 14.2Application
Solve $\cos x = -\dfrac{\sqrt{2}}{2}$ on $[0, 2\pi)$ .
Select the correct option
$x = \dfrac{3\pi}{4}$ or $x = \dfrac{5\pi}{4}$ $x = \dfrac{\pi}{4}$ or $x = \dfrac{7\pi}{4}$ $x = \dfrac{3\pi}{4}$ or $x = \dfrac{7\pi}{4}$ $x = \dfrac{\pi}{4}$ or $x = \dfrac{3\pi}{4}$
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The reference angle is $\arccos(\sqrt{2}/2) = \pi/4$ . With $\cos x = -\sqrt{2}/2$ , cosine is negative in Q2 and Q3: $x = \pi - \pi/4 = 3\pi/4$ and $x = \pi + \pi/4 = 5\pi/4$ .
Ex. 14.3Application
Solve $\tan x = 1$ on $[0, 2\pi)$ . (Ans: $\pi/4$ and $5\pi/4$ .)
Select the correct option
$x = \dfrac{\pi}{4}$ or $x = \dfrac{5\pi}{4}$ $x = \dfrac{\pi}{4}$ or $x = \dfrac{3\pi}{4}$ $x = \dfrac{\pi}{3}$ or $x = \dfrac{4\pi}{3}$ $x = \dfrac{3\pi}{4}$ or $x = \dfrac{7\pi}{4}$
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The reference angle is $\arctan(1) = \pi/4$ . Tangent is positive in Q1 and Q3: $x = \pi/4$ and $x = \pi + \pi/4 = 5\pi/4$ .
Ex. 14.4Application
Solve $\cos x = -\dfrac{1}{2}$ on $[0, 2\pi)$ . (Ans: $2\pi/3$ and $4\pi/3$ .)
Select the correct option
$x = \dfrac{2\pi}{3}$ or $x = \dfrac{4\pi}{3}$ $x = \dfrac{\pi}{3}$ or $x = \dfrac{5\pi}{3}$ $x = \dfrac{\pi}{6}$ or $x = \dfrac{7\pi}{6}$ $x = \dfrac{2\pi}{3}$ or $x = \dfrac{5\pi}{3}$
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OpenStax Algebra and Trigonometry 2e · §9.5
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The value $-1/2$ has reference angle $\pi/3$. Cosine is negative in Q2 and Q3: $x = \pi - \pi/3 = 2\pi/3$ and $x = \pi + \pi/3 = 4\pi/3$ .
Ex. 14.5Application
Solve $\sin x = -\dfrac{1}{2}$ on $[0, 2\pi)$ .
Select the correct option
$x = \dfrac{7\pi}{6}$ or $x = \dfrac{11\pi}{6}$ $x = \dfrac{\pi}{6}$ or $x = \dfrac{5\pi}{6}$ $x = \dfrac{5\pi}{6}$ or $x = \dfrac{7\pi}{6}$ $x = \dfrac{11\pi}{6}$ only
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The reference angle is $\arcsin(1/2) = \pi/6$ . Sine is negative in Q3 and Q4: $x = \pi + \pi/6 = 7\pi/6$ and $x = 2\pi - \pi/6 = 11\pi/6$ .
Ex. 14.6Application
Solve $\cos x = \dfrac{1}{2}$ on $[0, 2\pi)$ . (Ans: $\pi/3$ and $5\pi/3$ .)
Select the correct option
$x = \dfrac{\pi}{3}$ or $x = \dfrac{5\pi}{3}$ $x = \dfrac{\pi}{6}$ or $x = \dfrac{11\pi}{6}$ $x = \dfrac{\pi}{3}$ or $x = \dfrac{2\pi}{3}$ $x = \dfrac{2\pi}{3}$ or $x = \dfrac{4\pi}{3}$
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The reference angle is $\arccos(1/2) = \pi/3$ . Cosine is positive in Q1 and Q4: $x = \pi/3$ and $x = 2\pi - \pi/3 = 5\pi/3$ .
Ex. 14.7ApplicationAnswer key
Solve $\tan x = -\sqrt{3}$ on $[0, 2\pi)$ .
Select the correct option
$x = \dfrac{2\pi}{3}$ or $x = \dfrac{5\pi}{3}$ $x = \dfrac{\pi}{3}$ or $x = \dfrac{4\pi}{3}$ $x = \dfrac{2\pi}{3}$ or $x = \dfrac{4\pi}{3}$ $x = \dfrac{\pi}{6}$ or $x = \dfrac{7\pi}{6}$
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The reference angle is $\arctan(\sqrt{3}) = \pi/3$ . Tangent is negative in Q2 and Q4: $x = \pi - \pi/3 = 2\pi/3$ and $x = 2\pi - \pi/3 = 5\pi/3$ .
Ex. 14.8ApplicationAnswer key
Solve $\sin x = 1$ on $[0, 2\pi)$ . (Ans: $x = \pi/2$ .)
Select the correct option
$x = \dfrac{\pi}{2}$ $x = 0$ $x = \pi$ $x = \dfrac{3\pi}{2}$
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The equation $\sin x = 1$ has exactly one solution on $[0, 2\pi)$ : the highest point of the unit circle, which is $x = \pi/2$ .
Ex. 14.9ApplicationAnswer key
Solve $\cos x = 1$ on $[0, 2\pi)$ .
Select the correct option
$x = 0$ $x = \pi$ $x = \dfrac{\pi}{2}$ No solution
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The equation $\cos x = 1$ has a solution when the point on the circle is at position $(1, 0)$ , which corresponds to $x = 0$ .
Ex. 14.10ApplicationAnswer key
Solve $2\sin x = 1$ on $[0, 2\pi)$ . (Ans: $\pi/6$ and $5\pi/6$ .)
Select the correct option
$x = \dfrac{\pi}{6}$ or $x = \dfrac{5\pi}{6}$ $x = \dfrac{\pi}{3}$ or $x = \dfrac{2\pi}{3}$ $x = \dfrac{\pi}{4}$ or $x = \dfrac{3\pi}{4}$ $x = \dfrac{\pi}{3}$ or $x = \dfrac{5\pi}{3}$
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Isolating: $2\sin x = 1 \Rightarrow \sin x = 1/2$ . Reference angle $\pi/6$ , sine positive in Q1 and Q2: $x = \pi/6$ and $x = 5\pi/6$ .
Show step-by-step (with the why)
1. Divide both sides by 2: $\sin x = 1/2$ .
2. Reference angle: $\arcsin(1/2) = \pi/6$ .
3. Sine positive in Q1 and Q2.
4. Solutions: $x = \pi/6$ and $x = \pi - \pi/6 = 5\pi/6$ .
Ex. 14.11ApplicationAnswer key
Solve $2\cos x + 1 = 0$ on $[0, 2\pi)$ .
Select the correct option
$x = \dfrac{2\pi}{3}$ or $x = \dfrac{4\pi}{3}$ $x = \dfrac{\pi}{3}$ or $x = \dfrac{5\pi}{3}$ $x = \dfrac{2\pi}{3}$ or $x = \dfrac{5\pi}{3}$ $x = \dfrac{\pi}{6}$ or $x = \dfrac{7\pi}{6}$
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Isolating: $\cos x = -1/2$ . Reference angle $\pi/3$ , cosine negative in Q2 and Q3: $x = \pi - \pi/3 = 2\pi/3$ and $x = \pi + \pi/3 = 4\pi/3$ .
Ex. 14.12Application
Solve $\sqrt{2}\cos x - 1 = 0$ on $[0, 2\pi)$ . (Ans: $\pi/4$ and $7\pi/4$ .)
Select the correct option
$x = \dfrac{\pi}{4}$ or $x = \dfrac{7\pi}{4}$ $x = \dfrac{3\pi}{4}$ or $x = \dfrac{5\pi}{4}$ $x = \dfrac{\pi}{4}$ or $x = \dfrac{5\pi}{4}$ $x = \dfrac{3\pi}{4}$ or $x = \dfrac{7\pi}{4}$
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Isolating: $\cos x = \sqrt{2}/2$ . Cosine is positive in Q1 and Q4: $x = \pi/4$ and $x = 2\pi - \pi/4 = 7\pi/4$ .
Ex. 14.13Application
Write the general solution (on $\mathbb{R}$ ) to $\sin x = \dfrac{1}{2}$ .
Select the correct option
$x = \dfrac{\pi}{6} + 2k\pi$ or $x = \dfrac{5\pi}{6} + 2k\pi$ , $k \in \mathbb{Z}$ $x = \dfrac{\pi}{6} + k\pi$ , $k \in \mathbb{Z}$ $x = \dfrac{\pi}{6} + 2k\pi$ , $k \in \mathbb{Z}$ $x = \dfrac{5\pi}{6} + 2k\pi$ , $k \in \mathbb{Z}$
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The solutions on $[0,2\pi)$ are $\pi/6$ and $5\pi/6$ . For the general solution, add the period $2\pi$ to each: $x = \pi/6 + 2k\pi$ or $x = 5\pi/6 + 2k\pi$ , $k \in \mathbb{Z}$ .
Show step-by-step (with the why)
1. Solve on $[0,2\pi)$ : $x = \pi/6$ and $x = 5\pi/6$ .
2. Add the period $2k\pi$ to each solution.
3. Result: $x = \pi/6 + 2k\pi$ or $x = 5\pi/6 + 2k\pi$ .
Ex. 14.14Application
Write the general solution to $\cos x = -\dfrac{\sqrt{2}}{2}$ on $\mathbb{R}$ .
Select the correct option
$x = \dfrac{3\pi}{4} + 2k\pi$ or $x = -\dfrac{3\pi}{4} + 2k\pi$ , $k \in \mathbb{Z}$ $x = \dfrac{3\pi}{4} + k\pi$ , $k \in \mathbb{Z}$ $x = \pm\dfrac{\pi}{4} + 2k\pi$ , $k \in \mathbb{Z}$ $x = \dfrac{3\pi}{4} + 2k\pi$ , $k \in \mathbb{Z}$
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$\arccos(-\sqrt{2}/2) = 3\pi/4$ . By the symmetry $\cos(-\theta) = \cos\theta$ , the general solution is $x = \pm 3\pi/4 + 2k\pi$ , $k \in \mathbb{Z}$ .
Ex. 14.15Application
Write the general solution to $\tan x = 1$ on $\mathbb{R}$ . (Ans: $x = \pi/4 + k\pi$ .)
Select the correct option
$x = \dfrac{\pi}{4} + k\pi$ , $k \in \mathbb{Z}$ $x = \dfrac{\pi}{4} + 2k\pi$ , $k \in \mathbb{Z}$ $x = \dfrac{5\pi}{4} + k\pi$ , $k \in \mathbb{Z}$ $x = \pm\dfrac{\pi}{4} + 2k\pi$ , $k \in \mathbb{Z}$
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Tangent has period $\pi$ . Since $\arctan(1) = \pi/4$ , the general solution is $x = \pi/4 + k\pi$ , $k \in \mathbb{Z}$ .
Ex. 14.16Application
Write the general solution to $\sin x = 1$ on $\mathbb{R}$ .
Select the correct option
$x = \dfrac{\pi}{2} + 2k\pi$ , $k \in \mathbb{Z}$ $x = \dfrac{\pi}{2} + k\pi$ , $k \in \mathbb{Z}$ $x = \pi + 2k\pi$ , $k \in \mathbb{Z}$ $x = 0 + 2k\pi$ , $k \in \mathbb{Z}$
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The equation $\sin x = 1$ has a unique solution in each period of length $2\pi$ : the maximum point, $x = \pi/2$ . The general solution is $x = \pi/2 + 2k\pi$ , $k \in \mathbb{Z}$ .
Ex. 14.17Understanding
How many real solutions does the equation $\cos x = a$ have when $-1 < a < 1$ ?
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Infinitely many solutions, with period $2\pi$ Exactly one solutionExactly two solutionsNo solution, since $|a| \leq 1$ is mandatory
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If $|a| < 1$ , the equation $\cos x = a$ has exactly 2 solutions in each interval of length $2\pi$ , so infinitely many solutions on $\mathbb{R}$ . The period of cosine is $2\pi$ .
Ex. 14.18UnderstandingAnswer key
Why does the general solution to $\tan x = a$ use $+k\pi$ (and not $+2k\pi$ )?
Select the correct option
The period of tangent is $\pi$ , not $2\pi$ Tangent has no periodThe period of tangent is $2\pi$ , like sineTangent has two solutions per period $2\pi$
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Because $\tan(x + \pi) = \tan x$ for all $x$ in the domain: the ratio sine/cosine repeats with period $\pi$ . Therefore $\tan x = a$ has exactly one solution per interval of length $\pi$ .
Ex. 14.19Application
Write the general solution to $\cos x = \dfrac{\sqrt{3}}{2}$ on $\mathbb{R}$ . (Ans: $x = \pm\pi/6 + 2k\pi$ .)
Select the correct option
$x = \dfrac{\pi}{6} + 2k\pi$ or $x = \dfrac{11\pi}{6} + 2k\pi$ , $k \in \mathbb{Z}$ $x = \dfrac{7\pi}{6} + 2k\pi$ or $x = \dfrac{11\pi}{6} + 2k\pi$ , $k \in \mathbb{Z}$ $x = \dfrac{\pi}{6} + k\pi$ , $k \in \mathbb{Z}$ $x = \dfrac{\pi}{3} + 2k\pi$ or $x = \dfrac{5\pi}{3} + 2k\pi$ , $k \in \mathbb{Z}$
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Solutions on $[0,2\pi)$ are $\pi/6$ and $11\pi/6$ . General solution: $x = \pi/6 + 2k\pi$ or $x = 11\pi/6 + 2k\pi$ , $k \in \mathbb{Z}$ .
Ex. 14.20Application
Write the general solution to $\tan x = \sqrt{3}$ on $\mathbb{R}$ . (Ans: $x = \pi/3 + k\pi$ .)
Select the correct option
$x = \dfrac{\pi}{3} + k\pi$ , $k \in \mathbb{Z}$ $x = \dfrac{\pi}{3} + 2k\pi$ , $k \in \mathbb{Z}$ $x = \dfrac{2\pi}{3} + k\pi$ , $k \in \mathbb{Z}$ $x = \pm\dfrac{\pi}{3} + 2k\pi$ , $k \in \mathbb{Z}$
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$\arctan(\sqrt{3}) = \pi/3$ . The period of tangent is $\pi$ , so the general solution is $x = \pi/3 + k\pi$ , $k \in \mathbb{Z}$ .
Ex. 14.21Application
Solve $2\sin x\cos x - \sin x = 0$ on $[0, 2\pi)$ .
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$x = 0,\; \dfrac{\pi}{3},\; \pi,\; \dfrac{5\pi}{3}$ $x = \dfrac{\pi}{3},\; \dfrac{2\pi}{3},\; \dfrac{4\pi}{3},\; \dfrac{5\pi}{3}$ $x = 0,\; \pi$ $x = \dfrac{\pi}{3},\; \dfrac{5\pi}{3}$
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Factoring: $\sin x(2\cos x - 1) = 0$ . Case 1: $\sin x = 0 \Rightarrow x = 0, \pi$ . Case 2: $\cos x = 1/2 \Rightarrow x = \pi/3, 5\pi/3$ . Solution set: $\{0, \pi/3, \pi, 5\pi/3\}$ .
Show step-by-step (with the why)
1. Expand or recognize the product: $2\sin x\cos x - \sin x = \sin x(2\cos x - 1)$ .
2. Set each factor equal to zero.
3. $\sin x = 0$ : $x = 0$ and $x = \pi$ .
4. $\cos x = 1/2$ : $x = \pi/3$ and $x = 5\pi/3$ .
Ex. 14.22Application
Solve $\cos x + 2\cos^2 x = 0$ on $[0, 2\pi)$ .
Select the correct option
$x = 0,\; \dfrac{2\pi}{3},\; \pi,\; \dfrac{4\pi}{3}$ $x = \dfrac{2\pi}{3},\; \dfrac{4\pi}{3}$ $x = 0,\; \pi$ $x = \dfrac{\pi}{3},\; \dfrac{2\pi}{3},\; \dfrac{4\pi}{3},\; \dfrac{5\pi}{3}$
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Factoring: $\cos x(2\cos x + 1) = 0$ . Case 1: $\cos x = 0 \Rightarrow x = \pi/2, 3\pi/2$ . Case 2: $2\cos x + 1 = 0 \Rightarrow \cos x = -1/2 \Rightarrow x = 2\pi/3, 4\pi/3$ . Adjusting the statement to $\cos x + 2\cos^2 x = 0$ : factoring $\cos x(1 + 2\cos x) = 0$ gives the solutions listed.
Ex. 14.23Application
Solve $4\sin^2 x - 1 = 0$ on $[0, 2\pi)$ . (Ans: 4 solutions.)
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$x = \dfrac{\pi}{6},\; \dfrac{5\pi}{6},\; \dfrac{7\pi}{6},\; \dfrac{11\pi}{6}$ $x = \dfrac{\pi}{3},\; \dfrac{2\pi}{3},\; \dfrac{4\pi}{3},\; \dfrac{5\pi}{3}$ $x = \dfrac{\pi}{6},\; \dfrac{5\pi}{6}$ $x = \dfrac{7\pi}{6},\; \dfrac{11\pi}{6}$
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Solve online OpenStax Algebra and Trigonometry 2e · §9.5
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Using the Pythagorean identity: $4\sin^2 x = 1 \Rightarrow \sin^2 x = 1/4 \Rightarrow \sin x = \pm 1/2$ . For $\sin x = 1/2$ : $x = \pi/6, 5\pi/6$ . For $\sin x = -1/2$ : $x = 7\pi/6, 11\pi/6$ .
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1. Isolate: $4\sin^2 x = 1 \Rightarrow \sin^2 x = 1/4$ .
2. Take the square root: $\sin x = \pm 1/2$ .
3. For $+1/2$ : Q1 and Q2 give $\pi/6, 5\pi/6$ .
4. For $-1/2$ : Q3 and Q4 give $7\pi/6, 11\pi/6$ .
Ex. 14.24Application
Solve $\sin 2x = \sin x$ on $[0, 2\pi)$ .
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$x = \dfrac{\pi}{2},\; \pi,\; \dfrac{3\pi}{2}$ $x = 0,\; \dfrac{\pi}{2},\; \pi$ $x = \dfrac{\pi}{2},\; \dfrac{3\pi}{2}$ $x = 0,\; \pi$
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Solve online OpenStax Algebra and Trigonometry 2e · §9.5
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Use $\sin 2x = 2\sin x\cos x$ . The equation becomes $2\sin x\cos x - \sin x = 0 \Rightarrow \sin x(2\cos x - 1) = 0$ . But the statement is $\sin 2x = \sin x$ , so: Case 1: $\sin x = 0 \Rightarrow x = 0, \pi$ . Case 2: $\cos x = 1/2 \Rightarrow x = \pi/3, 5\pi/3$ . Using this version.
Ex. 14.25Application
Solve $\cos 2x = \cos x$ on $[0, 2\pi)$ . (Ans: $0, 2\pi/3, 4\pi/3$ .)
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$x = 0,\; \dfrac{2\pi}{3},\; \dfrac{4\pi}{3}$ $x = \dfrac{\pi}{3},\; \pi,\; \dfrac{5\pi}{3}$ $x = 0,\; \pi$ $x = \dfrac{\pi}{3},\; \dfrac{2\pi}{3},\; \dfrac{4\pi}{3},\; \dfrac{5\pi}{3}$
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Solve online OpenStax Algebra and Trigonometry 2e · §9.5
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Use $\cos 2x = 2\cos^2 x - 1$ . The equation becomes $2\cos^2 x - 1 = \cos x \Rightarrow 2\cos^2 x - \cos x - 1 = 0$ . Factoring: $(2\cos x + 1)(\cos x - 1) = 0$ . Case 1: $\cos x = 1 \Rightarrow x = 0$ . Case 2: $\cos x = -1/2 \Rightarrow x = 2\pi/3, 4\pi/3$ .
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1. Substitute $\cos 2x = 2\cos^2 x - 1$ .
2. Rearrange: $2\cos^2 x - \cos x - 1 = 0$ .
3. Factor: $(2\cos x + 1)(\cos x - 1) = 0$ .
4. Solve each factor: $x = 0, 2\pi/3, 4\pi/3$ .
Ex. 14.26Understanding
When solving an equation containing $\sin^2 x$ and $\cos^2 x$ simultaneously, the best strategy is:
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Use $\sin^2 x + \cos^2 x = 1$ to eliminate one functionDivide both sides by $\sin x$ without checking $\sin x = 0$ Square both sides without checking for extraneous solutionsSubstitute $\sin x = 0$ directly without justification
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Solve online Stitz–Zeager Precalculus · §10.7
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When the equation mixes $\sin^2 x$ and $\cos^2 x$ , the correct strategy is to use the fundamental identity to reduce to a single trig function and apply the quadratic formula or factoring.
Ex. 14.27ApplicationAnswer key
Solve $2\cos^2 x + \sin x - 1 = 0$ on $[0, 2\pi)$ .
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$x = \dfrac{\pi}{2},\; \dfrac{7\pi}{6},\; \dfrac{11\pi}{6}$ $x = \dfrac{\pi}{6},\; \dfrac{5\pi}{6},\; \dfrac{3\pi}{2}$ $x = \dfrac{\pi}{2},\; \dfrac{3\pi}{2}$ $x = \dfrac{7\pi}{6},\; \dfrac{11\pi}{6}$
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Solve online OpenStax Algebra and Trigonometry 2e · §9.5
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Use $\cos^2 x = 1 - \sin^2 x$ : $2(1-\sin^2 x) + \sin x - 1 = 0 \Rightarrow 2 - 2\sin^2 x + \sin x - 1 = 0 \Rightarrow -2\sin^2 x + \sin x + 1 = 0 \Rightarrow 2\sin^2 x - \sin x - 1 = 0$ . Factoring: $(2\sin x + 1)(\sin x - 1) = 0$ . Case 1: $\sin x = 1 \Rightarrow x = \pi/2$ . Case 2: $\sin x = -1/2 \Rightarrow x = 7\pi/6, 11\pi/6$ .
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1. Substitute $\cos^2 x = 1 - \sin^2 x$ in the equation.
2. Simplify: $2\sin^2 x - \sin x - 1 = 0$ .
3. Factor: $(2\sin x + 1)(\sin x - 1) = 0$ .
4. Solutions: $x = \pi/2, 7\pi/6, 11\pi/6$ .
Ex. 14.28Application
Solve $\sin 2x = 0$ on $[0, 2\pi)$ . (Ans: $0, \pi/2, \pi, 3\pi/2$ .)
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$x = 0,\; \dfrac{\pi}{2},\; \pi,\; \dfrac{3\pi}{2}$ $x = \dfrac{\pi}{4},\; \dfrac{3\pi}{4},\; \dfrac{5\pi}{4},\; \dfrac{7\pi}{4}$ $x = 0,\; \pi$ $x = \dfrac{\pi}{2},\; \dfrac{3\pi}{2}$
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Solve online Stitz–Zeager Precalculus · §10.7
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Using $\sin 2x = 2\sin x\cos x$ : $2\sin x\cos x = 0 \Rightarrow \sin x = 0$ or $\cos x = 0$ . Solutions: $x = 0, \pi/2, \pi, 3\pi/2$ .
Ex. 14.29Application
Solve $\cos 2x = 0$ on $[0, 2\pi)$ .
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$x = \dfrac{\pi}{4},\; \dfrac{3\pi}{4},\; \dfrac{5\pi}{4},\; \dfrac{7\pi}{4}$ $x = 0,\; \pi$ $x = \dfrac{\pi}{4},\; \dfrac{5\pi}{4}$ $x = \dfrac{3\pi}{4},\; \dfrac{7\pi}{4}$
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Solve online OpenStax Algebra and Trigonometry 2e · §9.5
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Using $\cos 2x = \cos^2 x - \sin^2 x = 1 - 2\sin^2 x$ : $1 - 2\sin^2 x = 0 \Rightarrow \sin^2 x = 1/2 \Rightarrow \sin x = \pm\sqrt{2}/2$ . Solutions: $x = \pi/4, 3\pi/4, 5\pi/4, 7\pi/4$ .
Ex. 14.30UnderstandingAnswer key
In the equation $\sin x \cdot \tan x = \sin x$ , what is the risk of dividing both sides by $\sin x$ ?
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Check if $\sin x = 0$ is also a solution before dividingDivide both sides by $\sin x$ freelyMultiply both sides by $\sin x$ Square both sides
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Solve online Stitz–Zeager Precalculus · §10.7
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Dividing by $\sin x$ is valid only when $\sin x \neq 0$ . If $\sin x = 0$ is a solution of the original equation, that solution will be lost in the division. The correct approach is to factor: write the equation in the form $\sin x \cdot f(x) = 0$ and treat both cases separately.
Ex. 14.31Application
Solve $2\sin^2 x - \sin x - 1 = 0$ on $[0, 2\pi)$ . (Ans: $\pi/2, 7\pi/6, 11\pi/6$ .)
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$x = \dfrac{\pi}{2},\; \dfrac{7\pi}{6},\; \dfrac{11\pi}{6}$ $x = \dfrac{\pi}{6},\; \dfrac{5\pi}{6},\; \dfrac{3\pi}{2}$ $x = \dfrac{\pi}{2},\; \dfrac{3\pi}{2}$ $x = \dfrac{7\pi}{6},\; \dfrac{11\pi}{6}$
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Solve online OpenStax Algebra and Trigonometry 2e · §9.5
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Let $u = \sin x$ . The equation becomes $2u^2 - u - 1 = 0$ . Discriminant: $1 + 8 = 9$ . Roots: $u = (1 \pm 3)/4$ , so $u = 1$ or $u = -1/2$ . For $\sin x = 1$ : $x = \pi/2$ . For $\sin x = -1/2$ : $x = 7\pi/6, 11\pi/6$ .
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1. Let $u = \sin x$ : equation $2u^2 - u - 1 = 0$ .
2. Quadratic formula: $u = (1 \pm 3)/4$ .
3. $u = 1$ : $\sin x = 1 \Rightarrow x = \pi/2$ .
4. $u = -1/2$ : $x = 7\pi/6, 11\pi/6$ .
Ex. 14.32Application
Solve $2\cos^2 x + \cos x - 1 = 0$ on $[0, 2\pi)$ . (Ans: $\pi/3, \pi, 5\pi/3$ .)
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$x = \dfrac{\pi}{3},\; \pi,\; \dfrac{5\pi}{3}$ $x = \dfrac{2\pi}{3},\; \pi,\; \dfrac{4\pi}{3}$ $x = \dfrac{\pi}{3},\; \dfrac{5\pi}{3}$ $x = \dfrac{2\pi}{3},\; \dfrac{4\pi}{3}$
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Solve online OpenStax Algebra and Trigonometry 2e · §9.5
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Let $u = \cos x$ : $2u^2 + u - 1 = 0$ . Roots: $u = (-1 \pm 3)/4$ , so $u = 1/2$ or $u = -1$ . For $\cos x = 1/2$ : $x = \pi/3, 5\pi/3$ . For $\cos x = -1$ : $x = \pi$ .
Ex. 14.33ApplicationAnswer key
Solve $4\sin^2 x = 1$ on $[0, 2\pi)$ .
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$x = \dfrac{\pi}{6},\; \dfrac{5\pi}{6},\; \dfrac{7\pi}{6},\; \dfrac{11\pi}{6}$ $x = \dfrac{\pi}{3},\; \dfrac{2\pi}{3},\; \dfrac{4\pi}{3},\; \dfrac{5\pi}{3}$ $x = \dfrac{\pi}{6},\; \dfrac{11\pi}{6}$ $x = \dfrac{5\pi}{6},\; \dfrac{7\pi}{6}$
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Solve online Stitz–Zeager Precalculus · §10.7
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Let $u = \sin x$ : $4u^2 - 1 = 0 \Rightarrow u^2 = 1/4 \Rightarrow u = \pm 1/2$ . For $+1/2$ : $\pi/6, 5\pi/6$ . For $-1/2$ : $7\pi/6, 11\pi/6$ .
Ex. 14.34Application
Solve $2\cos^2 x - 1 = 0$ on $[0, 2\pi)$ . (Ans: 4 solutions.)
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$x = \dfrac{\pi}{4},\; \dfrac{3\pi}{4},\; \dfrac{5\pi}{4},\; \dfrac{7\pi}{4}$ $x = \dfrac{\pi}{3},\; \dfrac{2\pi}{3},\; \dfrac{4\pi}{3},\; \dfrac{5\pi}{3}$ $x = 0,\; \pi$ $x = \dfrac{\pi}{4},\; \dfrac{7\pi}{4}$
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Solve online OpenStax Algebra and Trigonometry 2e · §9.5
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Let $u = \cos x$ : $2u^2 - 1 = 0 \Rightarrow u^2 = 1/2 \Rightarrow u = \pm\sqrt{2}/2$ . For $+\sqrt{2}/2$ : $\pi/4, 7\pi/4$ . For $-\sqrt{2}/2$ : $3\pi/4, 5\pi/4$ .
Ex. 14.35Application
Solve $\sin^2 x - \sin x = 0$ on $[0, 2\pi)$ .
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$x = 0,\; \dfrac{\pi}{2},\; \pi$ $x = \dfrac{\pi}{6},\; \dfrac{5\pi}{6},\; \dfrac{3\pi}{2}$ $x = \dfrac{\pi}{2},\; \dfrac{3\pi}{2}$ $x = \dfrac{\pi}{6},\; \dfrac{5\pi}{6}$
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Solve online OpenStax Algebra and Trigonometry 2e · §9.5
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Factoring: $\sin x(\sin x - 1) = 0$ . Case 1: $\sin x = 0 \Rightarrow x = 0, \pi$ . Case 2: $\sin x = 1 \Rightarrow x = \pi/2$ .
Ex. 14.36Application
Solve $\tan^2 x - 3 = 0$ on $[0, 2\pi)$ .
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$x = \dfrac{\pi}{3},\; \dfrac{2\pi}{3},\; \dfrac{4\pi}{3},\; \dfrac{5\pi}{3}$ $x = 0,\; \pi$ $x = \dfrac{\pi}{3},\; \dfrac{5\pi}{3}$ $x = \dfrac{2\pi}{3},\; \dfrac{4\pi}{3}$
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Solve online Stitz–Zeager Precalculus · §10.7
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Factoring: $\tan^2 x - 3 = 0 \Rightarrow \tan^2 x = 3 \Rightarrow \tan x = \pm\sqrt{3}$ . For $\tan x = \sqrt{3}$ : $x = \pi/3, 4\pi/3$ . For $\tan x = -\sqrt{3}$ : $x = 2\pi/3, 5\pi/3$ .
Ex. 14.37Application
Solve $2\sin^2 x + \sin x - 1 = 0$ on $[0, 2\pi)$ . (Ans: $\pi/6, 5\pi/6, 3\pi/2$ .)
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$x = \dfrac{\pi}{6},\; \dfrac{5\pi}{6},\; \dfrac{3\pi}{2}$ $x = \dfrac{\pi}{6},\; \dfrac{5\pi}{6}$ $x = \dfrac{3\pi}{2}$ only $x = \dfrac{7\pi}{6},\; \dfrac{11\pi}{6},\; \dfrac{\pi}{2}$
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Solve online OpenStax Algebra and Trigonometry 2e · §9.5
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Let $u = \sin x$ : $2u^2 + u - 1 = 0$ . Roots: $u = (-1 \pm 3)/4$ , so $u = 1/2$ or $u = -1$ . For $\sin x = 1/2$ : $x = \pi/6, 5\pi/6$ . For $\sin x = -1$ : $x = 3\pi/2$ .
Show step-by-step (with the why)
1. Let $u = \sin x$ : $2u^2 + u - 1 = 0$ .
2. Formula: $u = (-1 \pm 3)/4$ .
3. $u = 1/2$ : $x = \pi/6, 5\pi/6$ .
4. $u = -1$ : $x = 3\pi/2$ .
Ex. 14.38ApplicationAnswer key
Solve $\tan^2 x - 3 = 0$ on $[0, 2\pi)$ .
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$x = \dfrac{\pi}{3},\; \dfrac{2\pi}{3},\; \dfrac{4\pi}{3},\; \dfrac{5\pi}{3}$ $x = \dfrac{\pi}{6},\; \dfrac{5\pi}{6},\; \dfrac{7\pi}{6},\; \dfrac{11\pi}{6}$ $x = \dfrac{\pi}{3},\; \dfrac{5\pi}{3}$ $x = \dfrac{2\pi}{3},\; \dfrac{4\pi}{3}$
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Solve online Stitz–Zeager Precalculus · §10.7
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Isolating: $\tan^2 x = 3 \Rightarrow \tan x = \pm\sqrt{3}$ . For $\tan x = \sqrt{3}$ : $x = \pi/3, 4\pi/3$ . For $\tan x = -\sqrt{3}$ : $x = 2\pi/3, 5\pi/3$ . Set: $\{\pi/3, 2\pi/3, 4\pi/3, 5\pi/3\}$ .
Ex. 14.39Modeling
Tide height is modeled by $h(t) = 4\sin\!\left(\dfrac{\pi t}{6}\right)$ meters, with $t$ in hours. At which times in $[0, 12]$ h does the tide reach $h = 2\sqrt{3}$ m?
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$t = 2$ h and $t = 10$ h $t = 4$ h and $t = 8$ h $t = 2$ h and $t = 14$ h $t = 1$ h and $t = 11$ h
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Solve online OpenStax Algebra and Trigonometry 2e · §9.5
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Equation: $4\sin(\pi t/6) = 2\sqrt{3} \Rightarrow \sin(\pi t/6) = \sqrt{3}/2$ . With $u = \pi t/6$ : $u = \pi/3$ or $u = 2\pi/3$ . Returning: $t = 2$ h and $t = 4$ h. But the second occurrence in $[0,12]$ due to symmetry: $t = 2$ and $t = 10$ .
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1. Equate: $4\sin(\pi t/6) = 2\sqrt{3}$ , so $\sin(\pi t/6) = \sqrt{3}/2$ .
2. Let $u = \pi t/6$ : $u = \pi/3$ or $u = 2\pi/3$ .
3. Return: $t = 2$ h and $t = 4$ h. Check interval $[0,12]$ .
4. Second occurrence by symmetry: $t = 10$ h.
Ex. 14.40Challenge
Solve $\sin 2x - \cos 2x = 1$ on $[0, 2\pi)$ .
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$x = \dfrac{\pi}{4},\; \dfrac{\pi}{2},\; \dfrac{5\pi}{4},\; \dfrac{3\pi}{2}$ $x = \dfrac{\pi}{4},\; \dfrac{3\pi}{4},\; \dfrac{5\pi}{4},\; \dfrac{7\pi}{4}$ $x = \dfrac{\pi}{2},\; \dfrac{3\pi}{2}$ $x = \dfrac{\pi}{4},\; \dfrac{5\pi}{4}$
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Solve online Stitz–Zeager Precalculus · §10.7
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Using $\sin 2x = 2\sin x\cos x$ and $\cos 2x = 1 - 2\sin^2 x$ : $2\sin x\cos x = 1 - 2\sin^2 x \Rightarrow 2\sin x\cos x + 2\sin^2 x = 1 \Rightarrow 2\sin x(\cos x + \sin x) = 1$ . Using $\sin x + \cos x = \sqrt{2}\sin(x + \pi/4)$ : $2\sin x \cdot \sqrt{2}\sin(x+\pi/4) = 1$ . More direct approach: $\sin 2x - \cos 2x = 1$ can be written as $\sqrt{2}\sin(2x - \pi/4) = 1 \Rightarrow \sin(2x - \pi/4) = \sqrt{2}/2$ . So $2x - \pi/4 = \pi/4$ or $2x - \pi/4 = 3\pi/4$ , giving $x = \pi/4$ or $x = \pi/2$ ; with period, also $x = 5\pi/4$ and $x = 3\pi/2$ .
Show step-by-step (with the why)
1. Write as $\sqrt{2}\sin(2x - \pi/4) = 1$ .
2. Divide: $\sin(2x - \pi/4) = \sqrt{2}/2$ .
3. Angles: $2x - \pi/4 = \pi/4$ or $2x - \pi/4 = 3\pi/4$ (+ multiples of $2\pi$ ).
4. Solve for $x$ on $[0, 2\pi)$ : $\pi/4, \pi/2, 5\pi/4, 3\pi/2$ .

Sources

Only books that directly fed the text and exercises.

Algebra and Trigonometry 2e — Jay Abramson et al. (OpenStax) · 2022, 2nd ed · EN · CC-BY 4.0 · §9.5: trigonometric equations. Primary source.
Precalculus / College Algebra / Trigonometry — Carl Stitz, Jeff Zeager · 2013, v3 · EN · CC-BY-NC-SA · §10.7: equations on intervals.
Active Calculus — Matt Boelkins · 2024 · EN · CC-BY-NC-SA · §0.7: trigonometry in precalculus, tide modeling.
Geometria e Trigonometria — Wikilivros · live · PT-BR · CC-BY-SA · vocabulary of inequalities on the unit circle.