Table of Contents

Introduction
Trigonometric Functions and the Unit Circle
Solving Basic Trigonometric Equations
Linear Trigonometric Equations
Quadratic Trigonometric Equations
Points of Intersection
Trigonometric Inequalities
FAQs About AP Precalculus
- Q1: How Hard Is AP Precalculus?
- Q2: Does AP Precalculus Give College Credit?

3.10 Trigonometric Equations and Inequalities in AP Precalculus

Utkarsh
April 5, 2025

Table of Contents

Introduction
Trigonometric Functions and the Unit Circle
Solving Basic Trigonometric Equations
Linear Trigonometric Equations
Quadratic Trigonometric Equations
Points of Intersection
Trigonometric Inequalities
FAQs About AP Precalculus
- Q1: How Hard Is AP Precalculus?
- Q2: Does AP Precalculus Give College Credit?

Introduction

Trigonometric equations and inequalities are a fundamental component of AP Precalculus, specifically under Topic 3.10. These problems require you to manipulate trigonometric functions—sine, cosine, and tangent—while leveraging their periodic properties, symmetry, and the unit circle. The worksheets provided (A, B, C, and D by Bryan Passwater) present a variety of challenges: solving equations for exact values over the interval $[0, 2\pi)$ , finding points of intersection between functions, and determining intervals where inequalities hold. This blog post is a detailed, self-contained guide to mastering these concepts, complete with step-by-step explanations, solved examples and we have another detailed blog dedicated to the resources where we enlisted Top 10 books for AP Precalculus.

Trigonometric Functions and the Unit Circle

Before we tackle problem-solving, let’s establish the foundation. The sine, cosine, and tangent functions are defined on the unit circle, a circle with radius 1 centered at the origin. For an angle $\theta$ (in radians):

$\sin \theta$ represents the y-coordinate of the point on the circle.
$\cos \theta$ represents the x-coordinate.
$\tan \theta = \frac{\sin \theta}{\cos \theta}$ , which is undefined when $\cos \theta = 0$ .

The unit circle repeats every $2\pi$ radians, and within $[0, 2\pi)$ , each function exhibits distinct behavior:

Sine: Ranges from $-1$ to $1$ , positive in Quadrants I and II, negative in III and IV.
Cosine: Ranges from $-1$ to $1$ , positive in Quadrants I and IV, negative in II and III.
Tangent: Ranges from $-\infty$ to $\infty$ , positive in Quadrants I and III, negative in II and IV, undefined at $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ .

Key angles and their trigonometric values are critical to memorize:

$\sin \frac{\pi}{6} = \frac{1}{2}$ , $\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$ , $\tan \frac{\pi}{6} = \frac{\sqrt{3}}{3}$ .
$\sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}$ , $\cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}$ , $\tan \frac{\pi}{4} = 1$ .
$\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$ , $\cos \frac{\pi}{3} = \frac{1}{2}$ , $\tan \frac{\pi}{3} = \sqrt{3}$ .

Reference angles are used to find solutions across quadrants:

Quadrant I: $\theta = \text{reference angle}$ .
Quadrant II: $\theta = \pi - \text{reference angle}$ .
Quadrant III: $\theta = \pi + \text{reference angle}$ .
Quadrant IV: $\theta = 2\pi - \text{reference angle}$ .

Solving Basic Trigonometric Equations

Let’s begin with simple equations of the form $\sin x = k$ , $\cos x = k$ , or $\tan x = k$ .

Example 1: Solve the equation $\sin x = \frac{\sqrt{2}}{2}$ for all solutions in $[0, 2\pi)$ :

Step 1: Identify where sine equals the value. We need $\sin x = \frac{\sqrt{2}}{2}$ . From the unit circle, $\sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}$ , so the reference angle is $\frac{\pi}{4}$ .
Step 2: Determine quadrants. Sine is positive in Quadrants I and II.
Step 3: Find solutions.
- Quadrant I: $x = \frac{\pi}{4}$ .
- Quadrant II: $x = \pi - \frac{\pi}{4} = \frac{4\pi}{4} - \frac{\pi}{4} = \frac{3\pi}{4}$ .
Step 4: Verify interval. Both $\frac{\pi}{4}$ and $\frac{3\pi}{4}$ are within $[0, 2\pi)$ .

Solutions: $x = \frac{\pi}{4}, \frac{3\pi}{4}$ .

Example 2: Solve the equation $\tan x = -1$ for all solutions in $[0, 2\pi)$ :

Step 1: Reference angle. We need $\tan x = -1$ . Since $\tan \frac{\pi}{4} = 1$ , the reference angle is $\frac{\pi}{4}$ .
Step 2: Quadrants. Tangent is negative in Quadrants II and IV.
Step 3: Solutions.
- Quadrant II: $x = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$ .
- Quadrant IV: $x = 2\pi - \frac{\pi}{4} = \frac{8\pi}{4} - \frac{\pi}{4} = \frac{7\pi}{4}$ .
Step 4: Check. Both satisfy the interval.

Solutions: $x = \frac{3\pi}{4}, \frac{7\pi}{4}$ .

Download this free AP Precalculus worksheet with solutions for targeted practice and a better understanding of concepts

Start preparing effectively now!

Linear Trigonometric Equations

Equations with coefficients require algebraic manipulation to isolate the trigonometric function.

Example 3: Solve the equation $2 \cos x + 3 = 2$ for all solutions in $[0, 2\pi)$ :

Step 1: Isolate cosine.
- Start with $2 \cos x + 3 = 2$ .
- Subtract 3: $2 \cos x = 2 - 3 = -1$ .
- Divide by 2: $\cos x = -\frac{1}{2}$ .
Step 2: Reference angle. $\cos \frac{\pi}{3} = \frac{1}{2}$ , so the reference angle is $\frac{\pi}{3}$ .
Step 3: Quadrants. Cosine is negative in Quadrants II and III.
Step 4: Solutions.
- Quadrant II: $x = \pi - \frac{\pi}{3} = \frac{3\pi}{3} - \frac{\pi}{3} = \frac{2\pi}{3}$ .
- Quadrant III: $x = \pi + \frac{\pi}{3} = \frac{3\pi}{3} + \frac{\pi}{3} = \frac{4\pi}{3}$ .

Solutions: $x = \frac{2\pi}{3}, \frac{4\pi}{3}$ .

Example 4: Solve the equation $2 \sin x + \sqrt{3} = 0$ for all solutions in $[0, 2\pi)$ :

Step 1: Isolate sine.
- Start with $2 \sin x + \sqrt{3} = 0$ .
- Subtract $\sqrt{3}$ : $2 \sin x = -\sqrt{3}$ .
- Divide by 2: $\sin x = -\frac{\sqrt{3}}{2}$ .
Step 2: Reference angle. $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$ , reference angle is $\frac{\pi}{3}$ .
Step 3: Quadrants. Sine is negative in Quadrants III and IV.
Step 4: Solutions.
- Quadrant III: $x = \pi + \frac{\pi}{3} = \frac{4\pi}{3}$ .
- Quadrant IV: $x = 2\pi - \frac{\pi}{3} = \frac{6\pi}{3} - \frac{\pi}{3} = \frac{5\pi}{3}$ .

Solutions: $x = \frac{4\pi}{3}, \frac{5\pi}{3}$ .

Quadratic Trigonometric Equations

Equations with squared terms often resemble quadratics and may require factoring or the use of identities like $\sin^2 x + \cos^2 x = 1$ .

Example 5: Solve the equation $4 \sin^2 x + 2 = 5$ for all solutions in $[0, 2\pi)$ :

Step 1: Isolate the squared term.
- Start with $4 \sin^2 x + 2 = 5$ .
- Subtract 2: $4 \sin^2 x = 5 - 2 = 3$ .
- Divide by 4: $\sin^2 x = \frac{3}{4}$ .
Step 2: Solve for sine.
- $\sin x = \pm \sqrt{\frac{3}{4}} = \pm \frac{\sqrt{3}}{2}$ .
Step 3: Case 1: $\sin x = \frac{\sqrt{3}}{2}$ .
- Quadrants I and II.
- Quadrant I: $x = \frac{\pi}{3}$ .
- Quadrant II: $x = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$ .
Step 4: Case 2: $\sin x = -\frac{\sqrt{3}}{2}$ .
- Quadrants III and IV.
- Quadrant III: $x = \pi + \frac{\pi}{3} = \frac{4\pi}{3}$ .
- Quadrant IV: $x = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$ .

Solutions: $x = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}$ .

Example 6: Solve the equation $\cos^2 \theta = \cos \theta + 2$ for all solutions in $[0, 2\pi)$ :

Step 1: Form a quadratic.
- Start with $\cos^2 \theta = \cos \theta + 2$ .
- Move terms: $\cos^2 \theta - \cos \theta - 2 = 0$ .
Step 2: Factor.
- Let $u = \cos \theta$ : $u^2 - u - 2 = 0$ .
- $(u - 2)(u + 1) = 0$ .
- $u = 2$ or $u = -1$ .
Step 3: Solve for cosine.
- $\cos \theta = 2$ (impossible, since $-1 \leq \cos \theta \leq 1$ ).
- $\cos \theta = -1$ .
Step 4: Solutions.
- $\cos \theta = -1$ at $\theta = \pi$ .

Solution: $\theta = \pi$ .

Points of Intersection

These problems involve setting two functions equal and solving for $x$ .

Example 7: Find the $x$ -coordinates of the points of intersection of $f(x) = 5 - 3 \cos x$ and $g(x) = 5$ in $[0, 2\pi)$ :

Find $x$ -coordinates in $[0, 2\pi)$ where $f(x) = 5 - 3 \cos x$ and $g(x) = 5$ intersect:

Step 1: Set equal.
- $5 - 3 \cos x = 5$ .
Step 2: Solve.
- Subtract 5: $-3 \cos x = 0$ .
- Divide by -3: $\cos x = 0$ .
Step 3: Solutions.
- $\cos x = 0$ at $x = \frac{\pi}{2}, \frac{3\pi}{2}$ .

Solutions: $x = \frac{\pi}{2}, \frac{3\pi}{2}$ .

Example 8: Find the $x$ -coordinates of the points of intersection of $f(x) = \sin x$ and $g(x) = 2 \sin^2 x$ in $[0, 2\pi)$ :

Step 1: Set equal.
- $\sin x = 2 \sin^2 x$ .
Step 2: Rearrange.
- $2 \sin^2 x - \sin x = 0$ .
Step 3: Factor.
- $\sin x (2 \sin x - 1) = 0$ .
Step 4: Solve.
- $\sin x = 0$ : $x = 0, \pi$ (note: $2\pi$ is excluded by $< 2\pi$ ).
- .
  - Quadrant I: $x = \frac{\pi}{6}$ .
  - Quadrant II: $x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$ .

Solutions: $x = 0, \frac{\pi}{6}, \frac{5\pi}{6}, \pi$ .

Trigonometric Inequalities

Inequalities involve determining where a function satisfies a condition, often using the unit circle or sign analysis.

Example 9: Solve the inequality $\sin \theta \geq \frac{\sqrt{3}}{2}$ for all solutions in $[0, 2\pi)$ :

Step 1: Equality points.
- We need $\sin \theta \geq \frac{\sqrt{3}}{2}$ .
- $\sin \theta = \frac{\sqrt{3}}{2}$ at $\theta = \frac{\pi}{3}$ (Quadrant I), $\theta = \frac{2\pi}{3}$ (Quadrant II).
Step 2: Analyze sine’s behavior.
- Sine increases from $0$ to $1$ from $0$ to $\frac{\pi}{2}$ , peaks, then decreases to $-1$ at $\frac{3\pi}{2}$ , and rises to $0$ at $2\pi$ .
Step 3: Test intervals.
- $(0, \frac{\pi}{3})$ : $\sin \frac{\pi}{6} = 0.5 < 0.866$ , false.
- $(\frac{\pi}{3}, \frac{2\pi}{3})$ : $\sin \frac{\pi}{2} = 1 > 0.866$ , true.
- $(\frac{2\pi}{3}, 2\pi)$ : $\sin \pi = 0 < 0.866$ , false.
Step 4: Include endpoints.
- At $\frac{\pi}{3}$ and $\frac{2\pi}{3}$ , $\sin \theta = \frac{\sqrt{3}}{2}$ , satisfying $\geq$ .

Solution: $\theta \in \left[ \frac{\pi}{3}, \frac{2\pi}{3} \right]$ .

Example 10: Solve the inequality $(2 \cos \theta - 1)(\sin \theta + 1) < 0$ for all solutions in $[0, 2\pi]$ :

Solve for $[0, 2\pi]$ :

Step 1: Find zeros.
- We need $(2 \cos \theta - 1)(\sin \theta + 1) < 0$ .
- $2 \cos \theta - 1 = 0 \Rightarrow \cos \theta = \frac{1}{2} \Rightarrow \theta = \frac{\pi}{3}, \frac{5\pi}{3}$ .
- $\sin \theta + 1 = 0 \Rightarrow \sin \theta = -1 \Rightarrow \theta = \frac{3\pi}{2}$ .
Step 2: Critical points.
- $\frac{\pi}{3}, \frac{3\pi}{2}, \frac{5\pi}{3}$ (sort: $0, \frac{\pi}{3}, \frac{3\pi}{2}, \frac{5\pi}{3}, 2\pi$ ).
Step 3: Intervals.
- $(0, \frac{\pi}{3}), (\frac{\pi}{3}, \frac{3\pi}{2}), (\frac{3\pi}{2}, \frac{5\pi}{3}), (\frac{5\pi}{3}, 2\pi)$ .
Step 4: Sign analysis.
- Test $\theta = \frac{\pi}{6}$ in $(0, \frac{\pi}{3})$ : $(2 \cdot \frac{\sqrt{3}}{2} - 1)(\frac{1}{2} + 1) = (\sqrt{3} - 1)(1.5) > 0$ .
- Test $\theta = \pi$ in $(\frac{\pi}{3}, \frac{3\pi}{2})$ : $(2 \cdot (-1) - 1)(0 + 1) = (-3)(1) = -3 < 0$ .
- Test $\theta = \frac{7\pi}{4}$ in $(\frac{3\pi}{2}, \frac{5\pi}{3})$ : $(2 \cdot (-\frac{\sqrt{2}}{2}) - 1)(-\frac{\sqrt{2}}{2} + 1) \approx (-2.414)(0.293) < 0$ .
- Test $\theta = 2\pi$ in $(\frac{5\pi}{3}, 2\pi)$ : $(2 \cdot 1 - 1)(0 + 1) = (1)(1) > 0$ .
Step 5: Solution.
- Negative in $(\frac{\pi}{3}, \frac{3\pi}{2})$ and $(\frac{3\pi}{2}, \frac{5\pi}{3})$ .
- Union: $(\frac{\pi}{3}, \frac{5\pi}{3})$ (since $\frac{3\pi}{2}$ makes the product $0$ , excluded by $< 0$ ).

Solution: $\theta \in \left( \frac{\pi}{3}, \frac{5\pi}{3} \right)$ .

FAQs About AP Precalculus

As you explore Topic 3.10, you might have questions about AP Precalculus itself. Here are brief answers to two common ones:

Q1: How Hard Is AP Precalculus?

It’s moderately challenging if you’re solid on Algebra II and functions. Topic 3.10 adds trig complexity, but with practice (like the worksheets), it’s manageable—less intense than AP Calculus.

Q2: Does AP Precalculus Give College Credit?

Yes, many colleges offer credit or placement for a 3, 4, or 5 on the exam (3-4 credits), though policies vary. Check your target schools’ AP credit pages.

Need personalized guidance? Book a private online AP Precalculus tutor now!

Keep an eye on the official College Board web page for current information regarding the Exam!

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