7.5 Solving Trig Equations

Every trig equation follows the same arc: isolate the trig function, find the reference angle, check which quadrants give the right sign, list all solutions in the interval. The method changes — the arc doesn't.

Linear form: the trig function appears to the first power. Treat it like algebra — isolate the trig function, then find all angles that satisfy it on the given interval.

Example 1: 2cosθ − 3 = −5, on [0, 2π)

Add 3: 2cosθ = −2

Divide by 2: cosθ = −1

arccos(−1) = π. Cosine = −1 only at one point on [0, 2π).

θ = π

Example 2: cotθ + √3 = 0, on [0, 2π)

cotθ = −√3

cot = cos/sin. Reference angle: cot = √3 → θ_ref = π/6 (tan(π/6) = 1/√3, so cot = √3)

cot is negative in QII and QIV. But cot uses cotangent-style quadrants — negative where tan is negative: QII and QIV.

QII: π − π/6 = 5π/6. QIV: 2π − π/6 = 11π/6.

θ = 5π/6, 11π/6

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ASTC — which quadrants give positive values

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QII — Sin+
cos−, tan−

QI — All+
sin+, cos+, tan+

QIII — Tan+
sin−, cos−

QIV — Cos+
sin−, tan−

All Students Take Calculus. Once you know the reference angle and sign, use ASTC to find which quadrants work, then build the angle from the reference:

QI: θ = θ_ref · QII: π − θ_ref · QIII: π + θ_ref · QIV: 2π − θ_ref

Square root method: when you have trig² = k, take the square root of both sides. This gives ± so you get both positive and negative values — usually 4 solutions on [0, 2π).

Example 3: 2sin²θ − 1 = 0, on [0, 2π)

2sin²θ = 1 → sin²θ = 1/2

√(sin²θ) = ±√(1/2) → sinθ = ±1/√2 = ±√2/2

Reference angle: arcsin(√2/2) = π/4

sin+ → QI and QII: π/4, 3π/4. sin− → QIII and QIV: 5π/4, 7π/4

θ = π/4, 3π/4, 5π/4, 7π/4

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√(sin²θ) = |sinθ|, not sinθ — hence the ±

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When you take the square root of both sides of sin²θ = k, you get |sinθ| = √k, which means sinθ = +√k OR sinθ = −√k. Both cases must be solved. This is exactly like solving x² = 4 → x = ±2. You don't drop the negative. Dropping the ± is the most common mistake on square root method problems and will cost you half your solutions.

Square root solutions on the unit circle

Both ±√2/2 give solutions — positive (QI, QII) and negative (QIII, QIV)

Quadratic in form: looks like 2x² + x − 1 = 0 but with a trig function instead of x. Factor it (or use the quadratic formula), then solve each factor separately.

Example 4: 2cos²θ − cosθ − 1 = 0, on [0, 2π)

Let u = cosθ: 2u² − u − 1 = 0. Factor: (2u + 1)(u − 1) = 0

2cosθ + 1 = 0 → cosθ = −½. 2π/3 and 4π/3

cosθ − 1 = 0 → cosθ = 1 → θ = 0

θ = 0, 2π/3, 4π/3

Example 5: 2sin²θ − 5sinθ + 3 = 0, on [0, 2π)

Factor: (2sinθ − 3)(sinθ − 1) = 0

2sinθ = 3 → sinθ = 3/2. Impossible! |sinθ| ≤ 1. No solution from this factor.

sinθ = 1 → θ = π/2

θ = π/2

Example 6 (using identity): 3cosθ + 3 = 2sin²θ, on [0, 2π)

Substitute sin²θ = 1 − cos²θ: 3cosθ + 3 = 2(1 − cos²θ)

3cosθ + 3 = 2 − 2cos²θ → 2cos²θ + 3cosθ + 1 = 0

Factor: (2cosθ + 1)(cosθ + 1) = 0

cosθ = −½ → 2π/3, 4π/3. cosθ = −1 → π.

θ = 2π/3, π, 4π/3

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sinθ = 3/2 — why this has no solution

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The range of sine and cosine is always [−1, 1]. Any factor that produces sinθ = k where |k| > 1 has no real solution — just write "no solution" for that factor and continue with the others. This is different from "I made an error" — it just means that factor contributes nothing to the solution set. Same applies to cosθ = k where |k| > 1.

Multiple angles: when you have cos(2θ) or tan(3θ), solve for the inner argument first, then divide to get θ. You'll get more solutions because the period compresses.

Example 7: cos(2θ) = cosθ, on [0, 2π) — using double angle identity

Use cos(2θ) = 2cos²θ − 1: 2cos²θ − 1 = cosθ

2cos²θ − cosθ − 1 = 0 → (2cosθ + 1)(cosθ − 1) = 0

cosθ = −½ → 2π/3, 4π/3. cosθ = 1 → 0

θ = 0, 2π/3, 4π/3

Example 9: cos(2θ) = −½, on [0, 2π) — direct multiple angle

Let X = 2θ. Since θ ∈ [0, 2π), X = 2θ ∈ [0, 4π). Solve on the expanded interval.

cosX = −½. Reference angle = π/3. cos negative → QII and QIII.

First period: X = 2π/3, 4π/3. Second period: X = 2π/3 + 2π = 8π/3, X = 4π/3 + 2π = 10π/3.

Divide all by 2 to get θ: 2θ = 2π/3 → θ = π/3. 2θ = 4π/3 → θ = 2π/3. 2θ = 8π/3 → θ = 4π/3. 2θ = 10π/3 → θ = 5π/3.

θ = π/3, 2π/3, 4π/3, 5π/3

Multiple angle solution explorer — cos(nθ) = k

Multiple n n = 2

Target k k = −0.50

More solutions appear when n increases — the period compresses and more cycles fit in [0, 2π)

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Why multiple angles give more solutions

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cos(θ) has period 2π — one full cycle on [0, 2π), so usually 2 solutions. cos(2θ) has period π — two full cycles on [0, 2π), so usually 4 solutions. cos(3θ) has period 2π/3 — three full cycles, usually 6 solutions. The general rule: n·θ gives n times as many solutions as the base function. This is why you expand the interval to [0, 2nπ) before finding X = nθ solutions, then divide by n to get back to θ.

Example 10: √3 tan(3θ) + 1 = 0, on [0, 2π)

tan(3θ) = −1/√3. Let X = 3θ. X ∈ [0, 6π).

Reference angle: arctan(1/√3) = π/6. tan negative → QII and QIV.

Period of tan is π, so solutions: 5π/6, 11π/6, then +π each: 5π/6+π=11π/6... continuing for 3 full periods: X = 5π/6, 11π/6, 17π/6, 23π/6, 29π/6, 35π/6

Divide by 3: θ = 5π/18, 11π/18, 17π/18, 23π/18, 29π/18, 35π/18

θ = 5π/18, 11π/18, 17π/18, 23π/18, 29π/18, 35π/18

Using a calculator: when the answer isn't a special angle, use arcsin/arccos/arctan to get a decimal reference angle, then apply the quadrant logic to find all solutions.

Example 11: sinθ = 0.8, general solution in radians

θ_ref = arcsin(0.8) ≈ 0.9273 rad. sin positive → QI and QII.

QI: θ = 0.9273 + 2πk. QII: θ = π − 0.9273 = 2.2143 + 2πk.

θ ≈ 0.9273 + 2πk and θ ≈ 2.2143 + 2πk

Example 12: secθ = −4, general solution

secθ = 1/cosθ = −4 → cosθ = −1/4

θ_ref = arccos(1/4) ≈ 1.3181 rad. cos negative → QII and QIII.

QII: π − 1.3181 ≈ 1.8235. General: 1.8235 + 2πk

QIII: π + 1.3181 ≈ 4.4597. General: 4.4597 + 2πk

θ ≈ 1.8235 + 2πk and θ ≈ 4.4597 + 2πk

Calculator equation solver

k value k = 0.80

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Reciprocal functions on a calculator

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Your calculator doesn't have sec, csc, or cot buttons. Convert first:

secθ = k → cosθ = 1/k → use arccos(1/k)
cscθ = k → sinθ = 1/k → use arcsin(1/k)
cotθ = k → tanθ = 1/k → use arctan(1/k)

Example 12 above: secθ = −4 → cosθ = −1/4 → arccos(−1/4) ≈ 1.8235. Done.

Interactive unit circle solver. Drag the slider to set a target value. See where on the unit circle sinθ or cosθ hits that value — those are your solution angles.

k = 0.50

Solutions shown as highlighted points on the unit circle

5 methods at a glance

Linear: isolate trig function, find ref angle, use ASTC

Square root: trig² = k → trig = ±√k → both signs

Quadratic: factor like algebra, solve each factor

Multiple angles: expand interval, solve for nθ, divide by n

Calculator: get decimal ref angle, apply quadrant logic

Building angles from reference

QI: θ = θ_ref
QII: θ = π − θ_ref
QIII: θ = π + θ_ref
QIV: θ = 2π − θ_ref

General solutions:
sin/cos: + 2πk
tan/cot: + πk

Exam reminders

Always check if |k| > 1 for sin/cos — if so, no solution for that factor.

Square root method: NEVER drop the ±. Both signs give solutions.

Multiple angles: expand interval to [0, 2nπ) FIRST, then solve, then divide.

Never divide both sides by a trig function — factor instead or you lose solutions.

General solution: sin/cos add +2πk. tan/cot add +πk.

Calculator: reciprocal functions → flip k first, then apply arcsin/arccos/arctan.