Trigonometric Equations and Inequalities — AP Precalculus | The Complete Equation

AP PRECALCULUS — UNIT 3B · TRIGONOMETRIC & POLAR FUNCTIONS

3.10 — Trigonometric Equations and Inequalities

Notes — Solving Equations Involving Trig Functions

💡 Learning Objectives (3.10.A)

By the end of this lesson you will be able to:

Solve trigonometric equations on a restricted interval using reference angles
Write general solutions using +2πk (sine, cosine) or +πk (tangent)
Solve equations involving sin(bx), cos(bx), tan(bx) by adjusting the argument
Solve trigonometric inequalities by analyzing where the function is above or below a value

1. The Basic Strategy

To solve sin(θ) = c (or any other simple trig equation):

Identify the reference angle θ_ref using arcsin, arccos, or arctan applied to |c|
Determine which quadrants give the correct sign of c
Write the solutions in the requested interval (often [0, 2π))
If asked for ALL solutions, append +2πk (sine and cosine) or +πk (tangent)

2. Solving sin(θ) = c

📘 Example — Solve sin(θ) = √2/2 on [0, 2π)

Reference angle: arcsin(√2/2) = π/4
sin is positive in Quadrants I and II
Quadrant I: θ = π/4
Quadrant II: θ = π − π/4 = 3π/4
Solutions: θ ∈ {π/4, 3π/4}

📘 Example — General solution for sin(θ) = −1/2

Reference angle: arcsin(1/2) = π/6
sin is negative in Quadrants III and IV
Quadrant III: θ = π + π/6 = 7π/6
Quadrant IV: θ = 2π − π/6 = 11π/6
General: θ = 7π/6 + 2πk or θ = 11π/6 + 2πk, k ∈ ℤ

3. Solving cos(θ) = c

📘 Example — Solve cos(θ) = −1/2 on [0, 2π)

Reference angle: arccos(1/2) = π/3
cos is negative in Quadrants II and III
Quadrant II: θ = π − π/3 = 2π/3
Quadrant III: θ = π + π/3 = 4π/3
Solutions: θ ∈ {2π/3, 4π/3}

4. Solving tan(θ) = c

Tangent has period π, so it takes every real value EXACTLY ONCE in any interval of length π. Solutions repeat every π.

📘 Example — Solve tan(θ) = 1 on [0, 2π)

arctan(1) = π/4
Tan = 1 again at π/4 + π = 5π/4
Solutions: θ ∈ {π/4, 5π/4}
General: θ = π/4 + πk, k ∈ ℤ

5. Equations with Multiple Angles

For equations like sin(2x) = √3/2, treat 2x as the new variable u, solve for u, then divide by the multiplier — and PICK UP MORE SOLUTIONS along the way.

📘 Example — Solve sin(2x) = √3/2 on [0, 2π)

Let u = 2x. Solve sin(u) = √3/2.
u = π/3 or u = 2π/3 within [0, 2π)
Since 2x ∈ [0, 4π) for x ∈ [0, 2π), include MORE solutions: u ∈ {π/3, 2π/3, π/3 + 2π, 2π/3 + 2π} = {π/3, 2π/3, 7π/3, 8π/3}
Divide by 2: x ∈ {π/6, π/3, 7π/6, 4π/3}

⚠️ Common mistake

If x is in [0, 2π), then bx is in [0, 2πb) — so there are b times as many solutions as you'd expect. Forgetting this is the most common error in 3.10.

6. Equations Quadratic in Form

Equations like 2 sin²(x) − sin(x) − 1 = 0 are quadratic in sin(x). Substitute u = sin(x) and solve as a quadratic.

📘 Example — Quadratic in sin

2 sin²(x) − sin(x) − 1 = 0
Let u = sin(x): 2u² − u − 1 = 0
Factor: (2u + 1)(u − 1) = 0, so u = 1 or u = −1/2
Back-substitute: sin(x) = 1 ⇒ x = π/2; sin(x) = −1/2 ⇒ x = 7π/6 or 11π/6
Solutions on [0, 2π): {π/2, 7π/6, 11π/6}

7. Trig Inequalities

To solve a trig inequality like sin(x) > 1/2:

First find where sin(x) = 1/2 (the boundary points)
Test sample points between consecutive boundaries to determine the sign of the difference
Write the solution as the union of intervals where the inequality is satisfied

📘 Example — Solve sin(x) > 1/2 on [0, 2π)

Boundaries: sin(x) = 1/2 at x = π/6 and x = 5π/6
Test x = π/2 (between): sin(π/2) = 1 > 1/2 ✓
Test x = π (after 5π/6): sin(π) = 0 < 1/2 ✗
Test x = 0 (before π/6): sin(0) = 0 < 1/2 ✗
Solution: (π/6, 5π/6)

8. Summary

Use reference angles and quadrant signs to find solutions on [0, 2π)
Append +2πk for sine/cosine equations, +πk for tangent equations
For sin(bx), cos(bx), tan(bx), expect b times as many solutions in any base interval
Substitute u = trig(x) for quadratic-in-trig equations, solve algebraically, back-substitute
For inequalities, find boundaries (where equality holds), then test sample points

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