Solving Trig Equations with Identities

Section 7.1 — use identities to simplify, then solve like a regular equation

MAT 172 — Week 3
Solving a trig equation with identities is just algebra in disguise. Substitute the identity to get a single trig function, treat it like a regular equation, find the reference angle, then use ASTC to find all solutions in the requested interval.
The core strategy: Use Pythagorean, reciprocal, or quotient identities to rewrite the equation in terms of a single trig function. Then solve like algebra.
Step 1 — Simplify using identities
Look for sin²θ + cos²θ = 1 opportunities, or places where you can substitute tan = sin/cos, cot = cos/sin, sec = 1/cos, csc = 1/sin. Goal: one trig function, one equation.
Step 2 — Isolate the trig function
Get it in the form sin(θ) = k or cos(θ) = k or tan(θ) = k. This is just algebra — move terms, factor, divide.
Step 3 — Find the reference angle
Use arcsin, arccos, or arctan to find the reference angle. Use your special values table for exact answers (30°, 45°, 60°).
Step 4 — Find all solutions using ASTC
Determine which quadrants give the correct sign. For each valid quadrant, write the angle. If the problem asks for [0, 2π), list all solutions in that interval. If it says "general solution," add +2πn (for sin/cos) or +πn (for tan).
Identity substitution — which one to use Pythagorean identities sin²θ + cos²θ = 1 sin²θ = 1 − cos²θ cos²θ = 1 − sin²θ 1 + tan²θ = sec²θ cot²θ + 1 = csc²θ use when you see two functions Quotient identities tan θ = sin θ / cos θ cot θ = cos θ / sin θ use when you see tan or cot mixed with sin/cos Reciprocal identities csc θ = 1/sin θ sec θ = 1/cos θ cot θ = 1/tan θ use to eliminate csc, sec, cot
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Never divide both sides by a trig function
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If you have 2sin(θ)cos(θ) = 0 and you divide both sides by cos(θ), you lose the solutions where cos(θ) = 0. Instead, factor: 2sin(θ)cos(θ) = 0 → either sin(θ) = 0 OR cos(θ) = 0. Both possibilities give solutions. Dividing by a trig function is illegal because that function might equal zero at your solution — you'd be dividing by zero and losing answers.
✗ Never divide by sin(θ), cos(θ), or tan(θ) — always factor instead
Work through each step. Notice how every problem follows the same pattern — simplify with an identity, isolate, find reference angle, check quadrants.
Example 1: 2sin²θ − 1 = 0 on [0, 2π)
1
Isolate sin²θ: 2sin²θ = 1 → sin²θ = 1/2
2
Take square root: sin θ = ±√(1/2) = ±√2/2
3
Reference angle: arcsin(√2/2) = π/4 (45°)
4
sin positive → QI and QII: π/4 and 3π/4
5
sin negative → QIII and QIV: 5π/4 and 7π/4
θ = π/4, 3π/4, 5π/4, 7π/4
Example 2: 2cos²θ + cos θ − 1 = 0 on [0, 2π)
1
Recognize: this factors like a quadratic! Let u = cos θ: 2u² + u − 1 = 0
2
Factor: (2u − 1)(u + 1) = 0 → u = 1/2 or u = −1
3
cos θ = 1/2: reference angle = π/3. cos positive → QI and QIV: π/3 and 5π/3
4
cos θ = −1: arccos(−1) = π (only one solution)
θ = π/3, π, 5π/3
Example 3: sin²θ − cos²θ = 0 on [0, 2π)
1
Use Pythagorean: cos²θ = 1 − sin²θ. Substitute: sin²θ − (1 − sin²θ) = 0
2
Simplify: 2sin²θ − 1 = 0 → sin²θ = 1/2 → sin θ = ±√2/2
3
Reference angle: π/4. Both + and − → all four quadrants
θ = π/4, 3π/4, 5π/4, 7π/4
Example 4: tan θ − sin θ = 0 on [0, 2π)
1
Substitute tan θ = sin θ/cos θ: sin θ/cos θ − sin θ = 0
2
Factor sin θ: sin θ(1/cos θ − 1) = 0
3
sin θ = 0 → θ = 0, π
4
1/cos θ = 1 → cos θ = 1 → θ = 0 (already found)
θ = 0, π
General solutions vs interval solutions. When a problem says [0, 2π) — list only those solutions. When it says "all solutions" or "general solution" — add the period to capture infinitely many.
Period patterns — how solutions repeat sin and cos — period 2π If sin θ = k has solutions θ₁ and θ₂ in [0,2π): θ = θ₁ + 2πn, θ = θ₂ + 2πn Example: sin θ = √2/2 In [0,2π): θ = π/4 and θ = 3π/4 General: θ = π/4 + 2πn and θ = 3π/4 + 2πn where n is any integer n = 0 gives solutions in [0,2π) n = 1 adds 2π, n = −1 subtracts 2π... tan and cot — period π If tan θ = k has solution θ₁ in (−π/2,π/2): θ = θ₁ + πn Example: tan θ = 1 In [0,π): θ = π/4 In [0,2π): θ = π/4 AND 5π/4 General: θ = π/4 + πn One family covers both QI and QIII tan repeats every π, not 2π!
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How to count solutions on [0, 2π)
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For sin θ = k or cos θ = k: usually 2 solutions per period (one in each valid quadrant). Unless k = ±1 (only 1 solution) or |k| > 1 (no solution).

For tan θ = k: always exactly 1 solution per π interval, so 2 solutions in [0, 2π).

After factoring into multiple equations: solve each one separately, collect all solutions, remove duplicates, list in order from 0 to 2π.
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The "no solution" and "all θ" cases
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No solution: sin θ = 2 has no solution — sin is always between −1 and 1. cos θ = −3 has no solution for the same reason. Write "no solution" or ∅.

All real numbers: sin²θ + cos²θ = 1 is an identity — true for every θ. If you solve an equation and get a statement like this, the solution is all real numbers.
Drag the angle to see which trig equations it satisfies. Every point on the unit circle is a solution to some trig equation.
30°

At θ = 30°: sin = 0.500, cos = 0.866, tan = 0.577

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Key identities

sin²θ + cos²θ = 1
1 + tan²θ = sec²θ
cot²θ + 1 = csc²θ
tan θ = sin θ/cos θ
cot θ = cos θ/sin θ
General solutions:
sin/cos: + 2πn
tan/cot: + πn

ASTC reminder

QII: S
sin +
QI: A
all +
QIII: T
tan +
QIV: C
cos +

Exam reminders

Never divide both sides by a trig function — factor instead.
After getting trig function = k, always check both quadrants (unless k = ±1).
sin/cos equations: solutions repeat every 2π. tan/cot: every π.
If |k| > 1 for sin or cos equation: no solution.
Convert everything to one trig function before solving.