Solving Trigonometric Equations | FractionRush A-Level

Welcome to Solving Trigonometric Equations!

Solving trigonometric equations is a core skill in Cambridge A-Level Pure Mathematics 1. You will learn how to find all solutions of sin, cos and tan equations in a given interval, handle transformed arguments, form and solve quadratic trig equations, use Pythagorean identities to reduce mixed equations to a single ratio, and state general solutions valid for all integers.

sinθ = k → θ = arcsin(k) or π − arcsin(k) + 2nπ | cosθ = k → θ = ±arccos(k) + 2nπ | tanθ = k → θ = arctan(k) + nπ

Learning Objectives

Solve sinθ = k, cosθ = k, tanθ = k in [0°,360°] and [0,2π] using CAST
Adjust the interval and list all solutions for transformed arguments like sin(2θ) and sin(θ+30°)
Form and solve quadratic trig equations by substituting t = sinθ etc.
Use sin²θ + cos²θ = 1 and related identities to convert to one ratio
Write general solutions using the standard general solution formulas
Identify and reject inadmissible values from quadratics (|k| > 1)
Avoid the common error of dividing through by sinθ or cosθ
Apply step-by-step methods consistently under exam conditions

CAST Diagram

Which quadrant gives positive/negative sin, cos, tan?

Basic Equations

sinθ=k, cosθ=k, tanθ=k — principal value + all solutions

Transformed Arguments

sin(aθ+b) — adjust the interval, solve, back-substitute

Quadratic Trig

Let t = sinθ — factorise, check |t|≤1, find all angles

Identity Substitution

Use sin²+cos²=1 to reduce to one trig ratio

General Solutions

Formulas valid for all integer n — list solutions in a range

Learn 1 — Basic Trigonometric Equations

The CAST Diagram

The CAST diagram tells you which trig ratios are positive in each quadrant. This determines which angles to include as solutions.

S

sin +ve only

90°–180°

A

All +ve

0°–90°

T

tan +ve only

180°–270°

C

cos +ve only

270°–360°

Memory aid: CAST (going anti-clockwise from the 4th quadrant to the 1st)
A (1st quadrant, 0°–90°): All trig functions positive
S (2nd quadrant, 90°–180°): Sin positive only
T (3rd quadrant, 180°–270°): Tan positive only
C (4th quadrant, 270°–360°): Cos positive only

Solving sinθ = k in [0°, 360°]

Step 1: Find the principal value α = arcsin(k) (always in [−90°, 90°]).
Step 2: Use CAST to decide which quadrants give the correct sign of sin.
Step 3: Write both solutions using the quadrant rules for sin.

If sinθ = k (k > 0): θ = α and θ = 180° − α | If sinθ = k (k < 0): θ = 180° − α and θ = 360° + α

Example: Solve sinθ = 0.5 in [0°, 360°]
α = arcsin(0.5) = 30°
Since sin is positive: solutions in 1st and 2nd quadrants
θ = 30° and θ = 180° − 30° = 150°
Answer: θ = 30°, 150°

Example: Solve sinθ = −√3/2 in [0°, 360°]
α = arcsin(√3/2) = 60° (positive version)
Since sin is negative: solutions in 3rd and 4th quadrants
θ = 180° + 60° = 240° and θ = 360° − 60° = 300°
Answer: θ = 240°, 300°

Solving cosθ = k in [0°, 360°]

If cosθ = k (k > 0): θ = α and θ = 360° − α | If cosθ = k (k < 0): θ = 180° − α and 180° + α

Example: Solve cosθ = 1/2 in [0°, 360°]
α = arccos(1/2) = 60°
cos positive: 1st and 4th quadrants
θ = 60° and θ = 360° − 60° = 300°
Answer: θ = 60°, 300°

Example: Solve cosθ = −1/√2 in [0, 2π]
α = arccos(1/√2) = π/4
cos negative: 2nd and 3rd quadrants
θ = π − π/4 = 3π/4 and θ = π + π/4 = 5π/4
Answer: θ = 3π/4, 5π/4

Solving tanθ = k in [0°, 360°]

tanθ = k → θ = arctan(k) + 180°n — solutions always 180° apart

Example: Solve tanθ = 1 in [0°, 360°]
α = arctan(1) = 45°
tan positive: 1st and 3rd quadrants
θ = 45° and θ = 45° + 180° = 225°
Answer: θ = 45°, 225°

Example: Solve tanθ = −√3 in [0, 2π]
α = arctan(√3) = π/3
tan negative: 2nd and 4th quadrants
θ = π − π/3 = 2π/3 and θ = 2π − π/3 = 5π/3
Answer: θ = 2π/3, 5π/3

Always start by finding the positive principal value α = arcsin(|k|), arccos(|k|) or arctan(|k|), then use CAST to determine which quadrants apply based on the sign of k. In radians, replace 180° with π and 360° with 2π everywhere.

Learn 2 — Transformed Equations

Equations of the form sin(aθ) = k

When the argument is aθ instead of θ, you must adjust (expand) the interval. If θ ∈ [0°, 360°] and the argument is 2θ, then 2θ ∈ [0°, 720°]. Solve for 2θ in the expanded interval, then divide by 2 to get θ.

sin(aθ) = k in [0°,360°] → let u = aθ, solve sinu = k in [0°, 360°a], then θ = u/a

Example: Solve sin(2θ) = 1/2 in [0°, 360°]
Let u = 2θ. New interval: u ∈ [0°, 720°]
α = arcsin(1/2) = 30°
sin positive in 1st and 2nd quadrants. All solutions of sinu = 1/2 in [0°,720°]:
u = 30°, 150°, 360°+30° = 390°, 360°+150° = 510°
Divide by 2: θ = 15°, 75°, 195°, 255°
Answer: θ = 15°, 75°, 195°, 255°

Example: Solve cos(2θ) = √3/2 in [0°, 360°]
u = 2θ ∈ [0°, 720°]. α = arccos(√3/2) = 30°
cos positive: 1st and 4th quadrant solutions in each 360° cycle:
u = 30°, 330°, 390°, 690°
θ = 15°, 165°, 195°, 345°
Answer: θ = 15°, 165°, 195°, 345°

Equations of the form sin(θ + b) = k

Substitute u = θ + b. The interval for u is [0° + b, 360° + b]. Solve sinu = k in that shifted interval, then subtract b to find θ.

sin(θ + b) = k in [0°,360°] → u = θ+b ∈ [b, 360°+b], solve then θ = u − b

Example: Solve sin(θ + 30°) = 0.5 in [0°, 360°]
u = θ + 30°. Interval: u ∈ [30°, 390°]
α = arcsin(0.5) = 30°
sinu = 0.5 → u = 30° and u = 150° (within [30°,390°], also check 360°+30°=390° — boundary, include if ≤390°)
u = 30°: θ = 30° − 30° = 0° ✓ u = 150°: θ = 150° − 30° = 120° ✓ u = 390°: θ = 360° ✓
Answer: θ = 0°, 120°, 360°

Example: Solve cos(θ − 45°) = 1/√2 in [0°, 360°]
u = θ − 45°. Interval: u ∈ [−45°, 315°]
α = arccos(1/√2) = 45°. cos positive: u = 45° and u = −45°
u = 45°: θ = 90° ✓ u = −45°: θ = 0° ✓
Answer: θ = 0°, 90°

General method for sin(aθ + b) = k

Step 1: Let u = aθ + b
Step 2: Write the interval for u: u ∈ [a·0° + b, a·360° + b] = [b, 360°a + b]
Step 3: Find α = arcsin(|k|) (or arccos, arctan as appropriate)
Step 4: List ALL solutions of the equation in the expanded interval using CAST, adding 360° repeatedly
Step 5: Back-substitute: θ = (u − b)/a and check each value is in [0°, 360°]

Example: Solve tan(2θ − 60°) = 1 in [0°, 360°]
u = 2θ − 60°. Interval: u ∈ [−60°, 660°]
α = arctan(1) = 45°. tan positive: 1st and 3rd quadrants.
u = 45°, 225°, 405°, 585°
θ = (u + 60°)/2: θ = 52.5°, 142.5°, 232.5°, 322.5°
Answer: θ = 52.5°, 142.5°, 232.5°, 322.5°

The most common error in transformed equations is forgetting to adjust the interval. If you use u = 2θ but only look for solutions in [0°,360°], you will miss half the answers. Always write down the new interval for u before finding solutions.

Learn 3 — Quadratic Trigonometric Equations

Recognising a Quadratic in sinθ or cosθ

If an equation contains sin²θ and sinθ (with no cosθ), treat it as a quadratic in sinθ. Substitute t = sinθ, solve the quadratic in t, then find all angles for each valid value of t.

Equation in sin²θ and sinθ → let t = sinθ, factorise, check |t| ≤ 1, solve for θ

Step-by-Step: 2sin²θ − sinθ − 1 = 0

Step 1 — Substitute: Let t = sinθ. Equation becomes 2t² − t − 1 = 0
Step 2 — Factorise: (2t + 1)(t − 1) = 0
Step 3 — Solve for t: t = −1/2 or t = 1
Step 4 — Check |t| ≤ 1: Both −1/2 and 1 are valid ✓
Step 5 — Solve sinθ = −1/2 in [0°,360°]: α = 30°, sin negative → θ = 210°, 330°
Step 6 — Solve sinθ = 1 in [0°,360°]: θ = 90°
Answer: θ = 90°, 210°, 330°

Step-by-Step: 2cos²θ + cosθ − 1 = 0

Let t = cosθ → 2t² + t − 1 = 0
Factorise: (2t − 1)(t + 1) = 0
t = 1/2 or t = −1
Both valid: |1/2| ≤ 1, |−1| ≤ 1 ✓
cosθ = 1/2 → θ = 60°, 300°
cosθ = −1 → θ = 180°
Answer: θ = 60°, 180°, 300°

WARNING: Never Divide by sinθ or cosθ

Example of the fatal error — cos²θ = 3cosθ
✗ WRONG: Divide both sides by cosθ → cosθ = 3 → no solution. You LOST the solution cosθ = 0!

✓ CORRECT method: Rearrange to one side
cos²θ − 3cosθ = 0
cosθ(cosθ − 3) = 0
cosθ = 0 → θ = 90°, 270° cosθ = 3 → no solution (|3| > 1, rejected)
Answer: θ = 90°, 270°

Checking for Rejected Values

When solving quadratics you may get values of t outside [−1, 1]. For sinθ or cosθ these are impossible — mark them as "no solution" and move on. You must always state that the value is rejected.

Example: 3sin²θ + 2sinθ − 1 = 0
(3sinθ − 1)(sinθ + 1) = 0
sinθ = 1/3 ✓ (valid) → θ = arcsin(1/3) ≈ 19.5°, 160.5°
sinθ = −1 ✓ (valid) → θ = 270°
Answer: θ ≈ 19.5°, 160.5°, 270°

Quadratic in tanθ

tan has no restriction on its range, so all real values of tanθ are valid.

Example: 2tan²θ − tanθ − 3 = 0 in [0°, 360°]
Let t = tanθ: (2t − 3)(t + 1) = 0
tanθ = 3/2 → α = arctan(1.5) ≈ 56.3° → θ ≈ 56.3°, 236.3°
tanθ = −1 → α = 45° → θ = 135°, 315°
Answer: θ ≈ 56.3°, 135°, 236.3°, 315°

The substitution t = sinθ reduces the equation to a familiar quadratic. After factorising, always go back to θ by solving sinθ = each t value separately, and always list ALL solutions in the given interval for each.

Learn 4 — Identity Substitution

Key Identities for Solving Equations

sin²θ + cos²θ = 1 | sin²θ = 1 − cos²θ | cos²θ = 1 − sin²θ | tan²θ + 1 = sec²θ | tan θ = sinθ/cosθ

When an equation mixes sin²θ and cosθ (or cos²θ and sinθ), use sin²θ = 1 − cos²θ to convert everything to one ratio, then treat as a quadratic.

Method: Convert Mixed Equations to One Ratio

Step 1: Identify which trig ratio appears to a higher power (usually squared)
Step 2: Replace using sin²θ = 1 − cos²θ or cos²θ = 1 − sin²θ
Step 3: Simplify to get a quadratic in one ratio
Step 4: Solve the quadratic using substitution t = sinθ or t = cosθ
Step 5: Find all θ in the given interval, rejecting impossible values

Worked Example 1: 2sin²θ + 3cosθ − 3 = 0

Replace sin²θ with 1 − cos²θ:
2(1 − cos²θ) + 3cosθ − 3 = 0
2 − 2cos²θ + 3cosθ − 3 = 0
−2cos²θ + 3cosθ − 1 = 0
Multiply by −1: 2cos²θ − 3cosθ + 1 = 0
Let t = cosθ: (2t − 1)(t − 1) = 0
cosθ = 1/2 → θ = 60°, 300°
cosθ = 1 → θ = 0°, 360°
Answer (in (0°,360°)): θ = 60°, 300° (and 0°,360° if endpoints included)

Worked Example 2: 3cos²θ − sinθ − 2 = 0

Replace cos²θ with 1 − sin²θ:
3(1 − sin²θ) − sinθ − 2 = 0
3 − 3sin²θ − sinθ − 2 = 0
−3sin²θ − sinθ + 1 = 0
Multiply by −1: 3sin²θ + sinθ − 1 = 0
Using the formula: sinθ = (−1 ± √(1+12))/6 = (−1 ± √13)/6
sinθ = (−1 + 3.606)/6 ≈ 0.434 → θ ≈ 25.7°, 154.3°
sinθ = (−1 − 3.606)/6 ≈ −0.768 → θ ≈ 230.2°, 309.8°
Answer: θ ≈ 25.7°, 154.3°, 230.2°, 309.8°

Worked Example 3: tanθ = 2sinθ in [0°, 360°]

Replace tanθ with sinθ/cosθ:
sinθ/cosθ = 2sinθ
sinθ = 2sinθ · cosθ
sinθ − 2sinθcosθ = 0
sinθ(1 − 2cosθ) = 0
sinθ = 0 → θ = 0°, 180°, 360°
cosθ = 1/2 → θ = 60°, 300°
Answer: θ = 0°, 60°, 180°, 300°, 360°

The identity sin²θ + cos²θ = 1 is the most powerful tool for simplification. Whenever you see both sin²θ and cosθ (or cos²θ and sinθ) in the same equation, the substitution sin²θ = 1 − cos²θ will reduce it to a quadratic. This is a guaranteed Cambridge A-Level technique.

Learn 5 — General Solutions

What is a General Solution?

A general solution gives all possible solutions of a trig equation for every integer n, not just those in [0°,360°]. This is important for equations like sinθ = k where infinitely many angles satisfy the equation.

General Solution Formulas

sinθ = k → θ = (−1)ⁿ arcsin(k) + nπ (n ∈ ℤ)

cosθ = k → θ = ±arccos(k) + 2nπ (n ∈ ℤ)

tanθ = k → θ = arctan(k) + nπ (n ∈ ℤ)

In degrees, replace π with 180° and 2π with 360°:
sinθ = k → θ = (−1)ⁿ arcsin(k) + n·180°
cosθ = k → θ = ±arccos(k) + n·360°
tanθ = k → θ = arctan(k) + n·180°

Using General Solutions to List Values in a Range

Example: Solve sinθ = 1/2. Find all solutions in [0°, 720°].
General solution: θ = (−1)ⁿ · 30° + n·180°
n = 0: θ = 30°
n = 1: θ = (−1)¹·30° + 180° = −30° + 180° = 150°
n = 2: θ = (−1)²·30° + 360° = 30° + 360° = 390°
n = 3: θ = (−1)³·30° + 540° = −30° + 540° = 510°
n = 4: θ = 30° + 720° = 750° — outside range, stop
Answer: θ = 30°, 150°, 390°, 510°

Using the tan General Solution

Example: Solve tanθ = √3. Find all solutions in [−360°, 360°].
General solution: θ = 60° + n·180°
n = −2: θ = 60° − 360° = −300°
n = −1: θ = 60° − 180° = −120°
n = 0: θ = 60°
n = 1: θ = 60° + 180° = 240°
n = 2: θ = 60° + 360° = 420° — outside range
Answer: θ = −300°, −120°, 60°, 240°

Using the cos General Solution

Example: Solve cosθ = −1/2. Find all solutions in [0, 4π] (radians).
arccos(1/2) = π/3. cos negative: use π − π/3 = 2π/3 as the principal value in [π/2, π].
General solution: θ = ±2π/3 + 2nπ
n = 0: θ = 2π/3 and θ = −2π/3 (not in [0,4π])
n = 1: θ = 2π/3 + 2π = 8π/3 and θ = −2π/3 + 2π = 4π/3
n = 2: θ = 2π/3 + 4π = 14π/3 (outside) and θ = −2π/3 + 4π = 10π/3
Check: 10π/3 ≈ 10.47 and 4π ≈ 12.57 ✓
Answer: θ = 2π/3, 4π/3, 8π/3, 10π/3

For the sin general solution, the formula θ = (−1)ⁿ arcsin(k) + nπ alternates between the 1st-quadrant solution (n even) and the 2nd-quadrant solution (n odd). For tan, consecutive solutions are always exactly π apart. For cos, the ± captures both quadrant solutions simultaneously.

Worked Examples

Eight fully worked examples covering all key techniques for Cambridge A-Level 9709 trig equations.

Example 1 — sinθ = 1/2 in [0°, 360°]

Q: Solve sinθ = 1/2 in [0°, 360°].

M1: Principal value α = arcsin(1/2) = 30°. sin positive: 1st and 2nd quadrants.

A1: θ = 30° and θ = 180° − 30° = 150°. Answer: θ = 30°, 150° 2 marks

Example 2 — cos(2θ) = √3/2 in [0°, 360°]

Q: Solve cos(2θ) = √3/2 for 0° ≤ θ ≤ 360°.

M1: Let u = 2θ. New interval: u ∈ [0°, 720°]. α = arccos(√3/2) = 30°.

M1: cos positive: 1st and 4th quadrant solutions in each cycle. u = 30°, 330°, 390°, 690°.

A1: Divide by 2: θ = 15°, 165°, 195°, 345°. Answer: θ = 15°, 165°, 195°, 345° 3 marks

Example 3 — Quadratic in tanθ

Q: Solve 2tan²θ − 5tanθ + 2 = 0 in [0°, 360°].

M1: Let t = tanθ: (2t − 1)(t − 2) = 0. tanθ = 1/2 or tanθ = 2.

M1: tanθ = 1/2: α = arctan(0.5) ≈ 26.6°. Positive tan → θ ≈ 26.6°, 206.6°.

A1: tanθ = 2: α = arctan(2) ≈ 63.4°. Positive tan → θ ≈ 63.4°, 243.4°. Answer: θ ≈ 26.6°, 63.4°, 206.6°, 243.4° 4 marks

Example 4 — Identity Substitution: 2sin²θ + cosθ = 1

Q: Solve 2sin²θ + cosθ = 1 for 0° ≤ θ ≤ 360°.

M1: Replace sin²θ = 1 − cos²θ: 2(1 − cos²θ) + cosθ = 1 → 2 − 2cos²θ + cosθ = 1.

M1: 2cos²θ − cosθ − 1 = 0. Let t = cosθ: (2t + 1)(t − 1) = 0. cosθ = −1/2 or cosθ = 1.

A1: cosθ = −1/2 → θ = 120°, 240°. cosθ = 1 → θ = 0°, 360°. Answer: θ = 0°, 120°, 240°, 360° 4 marks

Example 5 — sin(θ − 45°) = √2/2 in [0°, 360°]

Q: Solve sin(θ − 45°) = √2/2 for 0° ≤ θ ≤ 360°.

M1: Let u = θ − 45°. Interval: u ∈ [−45°, 315°]. α = arcsin(√2/2) = 45°.

M1: sin positive: u = 45° and u = 135°. Check u = −45° is not in interval for this positive k.

A1: θ = 45° + 45° = 90° and θ = 135° + 45° = 180°. Answer: θ = 90°, 180° 3 marks

Example 6 — General Solution: sinθ = −√3/2

Q: Write the general solution of sinθ = −√3/2.

M1: Principal value arcsin(√3/2) = π/3 = 60°. Since sin is negative, the principal value is −π/3.

M1: General solution formula: θ = (−1)ⁿ(−π/3) + nπ

A1: Alternatively: θ = nπ + (−1)ⁿ⁺¹(π/3), n ∈ ℤ. This gives θ = π + π/3, 2π − π/3, ... i.e. 4π/3, 5π/3 in [0,2π]. 3 marks

Example 7 — Do Not Divide by cosθ

Q: Solve sinθ = 3cosθ in [0°, 360°].

M1: Divide both sides by cosθ (valid here since we check cosθ=0 separately): tanθ = 3. Note cosθ=0 gives sinθ=0, contradiction, so cosθ≠0.

M1: α = arctan(3) ≈ 71.6°. tan positive → 1st and 3rd quadrants.

A1: θ ≈ 71.6° and θ ≈ 251.6°. Answer: θ ≈ 71.6°, 251.6° 3 marks

Example 8 — Combined: sin(2θ) + cos(2θ) = 0 in [0, 2π]

Q: Solve sin(2θ) + cos(2θ) = 0 in [0, 2π].

M1: Divide by cos(2θ): tan(2θ) = −1. Let u = 2θ ∈ [0, 4π]. α = arctan(1) = π/4.

M1: tan negative: 2nd and 4th quadrants. u = 3π/4, 7π/4, 3π/4+2π = 11π/4, 7π/4+2π = 15π/4.

A1: θ = u/2: θ = 3π/8, 7π/8, 11π/8, 15π/8. Answer: θ = 3π/8, 7π/8, 11π/8, 15π/8 4 marks

Common Mistakes in Solving Trig Equations

These are the errors examiners see most frequently. Study each one carefully.

Mistake 1 — Giving only the principal value (missing the second solution)

✗ sinθ = 0.5 → θ = 30° only

✓ sinθ = 0.5 → θ = 30° AND θ = 150°. Always use CAST to find the second solution in the interval.

Mistake 2 — Forgetting to adjust the interval for transformed equations

✗ Solving sin(2θ) = k by looking for solutions in [0°, 360°] instead of [0°, 720°]

✓ For sin(2θ), let u = 2θ and solve in [0°, 720°]. You will find up to 4 solutions — halving them gives all values of θ in [0°, 360°].

Mistake 3 — Dividing through by sinθ or cosθ

✗ cos²θ = cosθ → divide by cosθ → cosθ = 1 → θ = 0°. Missed θ = 90°, 270°!

✓ ALWAYS factorise: cos²θ − cosθ = 0 → cosθ(cosθ − 1) = 0. Both factors give solutions.

Mistake 4 — Not rejecting values where |t| > 1

✗ Getting sinθ = 3 from a quadratic and trying to find an angle for it

✓ Since −1 ≤ sinθ ≤ 1, any value of t with |t| > 1 is impossible. Write "no solution from sinθ = 3" and move on.

Mistake 5 — Wrong quadrant for negative values

✗ sinθ = −0.5 → θ = −30° (principal value only, which is not in [0°,360°])

✓ sinθ = −0.5: sin is negative in 3rd and 4th quadrants. α = 30°, so θ = 180° + 30° = 210° and θ = 360° − 30° = 330°.

Mistake 6 — Forgetting ± in the back-substitution for shifted interval

✗ sin(θ + 30°) = k: after solving u = θ + 30°, forgetting to subtract 30° from every solution

✓ Every u value must be back-substituted: θ = u − 30°. Check each resulting θ lies in the original interval [0°, 360°].

Mistake 7 — Taking the square root and forgetting ±

✗ cos²θ = 3/4 → cosθ = √3/2 only

✓ cos²θ = 3/4 → cosθ = ±√3/2. Solve both cosθ = √3/2 (θ = 30°, 330°) and cosθ = −√3/2 (θ = 150°, 210°). All four solutions must be found.

Mistake 8 — Applying identity incorrectly in substitution

✗ In 3sin²θ + cosθ = 2, replacing cos²θ with 1 − sin²θ (wrong identity; cosθ is not squared here)

✓ Since sin²θ appears, replace sin²θ = 1 − cos²θ to get everything in terms of cosθ. Only substitute the squared identity for the squared term.

Mistake 9 — Rounding principal value before finding all solutions

✗ Computing arcsin(0.7) ≈ 44.4° and rounding to 44° before finding the second solution 136°

✓ Use the exact principal value through the quadrant step. Only round the final answers — e.g., θ = 44.4° and θ = 180° − 44.4° = 135.6°.

Key Formulas — Solving Trigonometric Equations

Equation Type	Formula / Rule	Notes
sinθ = k (k>0)	θ = α and θ = 180°−α (α = arcsin k)	1st and 2nd quadrant
sinθ = k (k<0)	θ = 180°−α and θ = 360°+α	3rd and 4th quadrant
cosθ = k (k>0)	θ = α and θ = 360°−α (α = arccos k)	1st and 4th quadrant
cosθ = k (k<0)	θ = 180°−α and θ = 180°+α	2nd and 3rd quadrant
tanθ = k (any sign)	θ = α + 180°n, n = 0,1 in [0°,360°]	solutions 180° apart
CAST — All	1st quadrant: all positive	0° to 90°
CAST — Sin	2nd quadrant: sin positive	90° to 180°
CAST — Tan	3rd quadrant: tan positive	180° to 270°
CAST — Cos	4th quadrant: cos positive	270° to 360°
Adjusted interval	sin(aθ+b)=k: u=aθ+b ∈ [b, 360a+b]	expand then solve
Key identity 1	sin²θ + cos²θ = 1	convert between sin and cos
Key identity 2	sin²θ = 1 − cos²θ	replace squared term
Key identity 3	cos²θ = 1 − sin²θ	replace squared term
Key identity 4	tan θ = sinθ/cosθ	reduce tanθ equations
Key identity 5	1 + tan²θ = sec²θ	used when secθ appears
General: sinθ = k	θ = (−1)ⁿ arcsin(k) + nπ, n∈ℤ	all solutions
General: cosθ = k	θ = ±arccos(k) + 2nπ, n∈ℤ	all solutions
General: tanθ = k	θ = arctan(k) + nπ, n∈ℤ	all solutions
Reject if	\|sinθ\| > 1 or \|cosθ\| > 1	no solution — state "rejected"

Proof Bank

Two key proofs that underpin the solution methods for trig equations.

Proof 1 — Why dividing by sinθ loses solutions

Consider the equation sinθ · (sinθ − 1) = 0.

There are two factors: sinθ = 0 gives θ = 0°, 180°, 360°; and sinθ = 1 gives θ = 90°.

Now suppose someone "divides both sides by sinθ". This assumes sinθ ≠ 0 — but sinθ = 0 IS a solution. Dividing by zero (or a factor that can equal zero) eliminates those solutions from the equation entirely.

Formal argument: Dividing both sides of A·B = 0 by A is only valid if A ≠ 0. If A could be zero, then A = 0 itself is a solution branch that must be kept. The correct approach is always to factorise and apply the zero-product property: if A·B = 0 then A = 0 or B = 0 (or both).

Consequence: In any trig equation, NEVER cancel sinθ or cosθ from both sides of an equation. Move it to one side and factorise.

Proof 2 — Derivation of the general solution sinθ = k from graph periodicity

The graph y = sinθ has period 2π: it repeats every 2π radians.

For a given value k ∈ [−1,1], the equation sinθ = k has exactly two solutions per 2π period.

Let α = arcsin(k) be the principal value (α ∈ [−π/2, π/2]).

First solution in [0, 2π]: θ₁ = α (if α ≥ 0) or θ₁ = 2π + α (if α < 0).
Second solution in [0, 2π]: By the symmetry of the sine graph about θ = π/2: θ₂ = π − α.

These repeat every 2π, so the complete solution set is:
θ = α + 2nπ or θ = π − α + 2nπ, for all n ∈ ℤ.

These two families can be combined into one formula using (−1)ⁿ:
When n is even: (−1)ⁿα + nπ = α + nπ → for n = 0, 2, 4,... gives α, α+2π, α+4π,... ✓
When n is odd: (−1)ⁿα + nπ = −α + nπ → for n = 1: −α + π = π − α ✓; n = 3: −α + 3π = (π−α) + 2π ✓

Therefore: sinθ = k → θ = (−1)ⁿ arcsin(k) + nπ, n ∈ ℤ Q.E.D.

Trig Equation Visualiser

Select a trig function and drag the k slider to see the horizontal line y = k. All intersection points with the curve in [0, 2π] are highlighted — these are the solutions.

Function:

k = 0.50

Move the slider to see solutions highlighted on the curve.

Exercise 1 — Basic Trig Equations in [0°, 360°] (10 Questions)

Give all solutions in [0°,360°]. For multiple answers, enter them separated by commas in ascending order (e.g. 30,150). Round to 1 decimal place where needed.

Exercise 2 — Transformed Equations (10 Questions)

Enter the number of solutions found, or the smallest positive solution in degrees (as stated). Round to 1 dp.

Exercise 3 — Quadratic Trig Equations (10 Questions)

Enter the number of valid solutions in [0°,360°] unless otherwise stated.

Exercise 4 — Identity Substitution (10 Questions)

Enter the number of solutions or the specific angle as requested.

Exercise 5 — General Solutions (10 Questions)

Enter the number of solutions in the specified interval, or the value as requested.

Practice — 30 Mixed Questions

Mixed trig equation problems. Enter numerical answers as prompted. Get 100% for confetti!

Challenge — 15 Harder Questions

More demanding questions requiring multiple steps. Enter answers as instructed.

Exam Style Questions

8 Cambridge-style questions with mark schemes. Attempt each before revealing the solution.

Q1 [3 marks]

Solve the equation sinθ = −√3/2 for 0° ≤ θ ≤ 360°.

α = arcsin(√3/2) = 60° [B1]
sin negative → 3rd and 4th quadrants [M1]
θ = 180° + 60° = 240° and θ = 360° − 60° = 300° [A1]

Q2 [4 marks]

Solve cos(2θ) = −1/2 for 0° ≤ θ ≤ 360°.

Let u = 2θ ∈ [0°, 720°] [M1 correct interval]
α = arccos(1/2) = 60°. cos negative → u = 120°, 240°, 480°, 600° [M1 A1 four solutions]
θ = 60°, 120°, 240°, 300° [A1 halving correctly]

Q3 [4 marks]

Solve 2sin²θ − 3sinθ + 1 = 0 for 0° ≤ θ ≤ 360°.

Let t = sinθ: 2t² − 3t + 1 = 0 → (2t − 1)(t − 1) = 0 [M1 factorise]
sinθ = 1/2: θ = 30°, 150° [A1]
sinθ = 1: θ = 90° [A1]
All three values correct [A1]

Q4 [5 marks]

Solve 3cos²θ + cosθ − 2 = 0 for 0 ≤ θ ≤ 2π. Give answers in exact radians.

(3cosθ − 2)(cosθ + 1) = 0 [M1 factorise]
cosθ = 2/3: θ = arccos(2/3) ≈ 0.841 and θ = 2π − 0.841 ≈ 5.442 [M1 A1]
cosθ = −1: θ = π [B1]
All three answers stated [A1]

Q5 [5 marks]

Solve 3sin²θ − 2cosθ − 2 = 0 for 0° ≤ θ ≤ 360°.

Replace sin²θ = 1 − cos²θ: 3(1 − cos²θ) − 2cosθ − 2 = 0 [M1]
3 − 3cos²θ − 2cosθ − 2 = 0 → 3cos²θ + 2cosθ − 1 = 0 [M1]
(3cosθ − 1)(cosθ + 1) = 0 [M1 factorise]
cosθ = 1/3 → θ = 70.5°, 289.5° [A1]
cosθ = −1 → θ = 180° [A1]

Q6 [4 marks]

Solve sin(θ + 60°) = cos60° for 0° ≤ θ ≤ 360°.

cos60° = 0.5. Let u = θ + 60°, u ∈ [60°, 420°] [M1]
sinu = 0.5, α = 30°. Positive sin → u = 30° (not in interval), u = 150°, u = 30°+360°=390° [M1 A1]
θ = 150° − 60° = 90° and θ = 390° − 60° = 330° [A1]

Q7 [4 marks]

Given that tanθ = sinθ/cosθ, solve tanθ = 2cosθ for 0 ≤ θ ≤ 2π.

sinθ/cosθ = 2cosθ → sinθ = 2cos²θ [M1]
sinθ = 2(1 − sin²θ) → 2sin²θ + sinθ − 2 = 0 [M1]
sinθ = (−1 ± √17)/4. Only sinθ = (−1+√17)/4 ≈ 0.780 is valid (|other| > 1 rejected) [M1]
θ = arcsin(0.780) ≈ 0.894 rad and θ = π − 0.894 ≈ 2.248 rad [A1]

Q8 [5 marks]

(a) Show that cos²θ − sin²θ = 2cos²θ − 1. [1 mark]
(b) Hence solve cos²θ − sin²θ = cosθ for 0° ≤ θ ≤ 360°. [4 marks]

(a) cos²θ − sin²θ = cos²θ − (1 − cos²θ) = 2cos²θ − 1 [B1]
(b) 2cos²θ − 1 = cosθ → 2cos²θ − cosθ − 1 = 0 [M1]
(2cosθ + 1)(cosθ − 1) = 0 [M1]
cosθ = −1/2 → θ = 120°, 240° [A1]
cosθ = 1 → θ = 0°, 360° [A1]

Past Paper Questions

5 realistic past-paper style questions based on Cambridge A-Level 9709 Pure 1 trig equation topics. Click to reveal full solutions.

Past Paper Q1 — 9709/11 style [4 marks]

Solve the equation 4sin²x − 1 = 0 for 0° ≤ x ≤ 360°.

4sin²x = 1 → sin²x = 1/4 → sinx = ±1/2 [M1 taking square root with ±]
sinx = 1/2 → x = 30°, 150° [A1]
sinx = −1/2 → x = 210°, 330° [A1]
All four answers: x = 30°, 150°, 210°, 330° [A1]

Past Paper Q2 — 9709/12 style [5 marks]

Solve the equation 2sin(2θ) − 1 = 0 for 0 ≤ θ ≤ 2π. Give exact answers in terms of π.

sin(2θ) = 1/2. Let u = 2θ ∈ [0, 4π] [M1 correct interval]
α = arcsin(1/2) = π/6. sin positive → u = π/6, 5π/6, π/6+2π = 13π/6, 5π/6+2π = 17π/6 [M1 A1 four solutions]
θ = π/12, 5π/12, 13π/12, 17π/12 [A1 A1 all correct]
Answer: θ = π/12, 5π/12, 13π/12, 17π/12

Past Paper Q3 — 9709/13 style [5 marks]

Solve the equation 2sin²θ + sinθcosθ = 0 for 0° ≤ θ ≤ 360°.

Factorise: sinθ(2sinθ + cosθ) = 0 [M1 factorise sinθ]
sinθ = 0 → θ = 0°, 180°, 360° [A1]
2sinθ + cosθ = 0 → 2sinθ = −cosθ → tanθ = −1/2 [M1 divide by cosθ; valid since if cosθ=0 then sinθ=0 which contradicts sinθ≠0 here]
α = arctan(1/2) ≈ 26.6°. tan negative → θ = 180° − 26.6° = 153.4° and θ = 360° − 26.6° = 333.4° [A1 A1]
Answer: θ = 0°, 153.4°, 180°, 333.4°, 360°

Past Paper Q4 — 9709/11 style [6 marks]

Solve the equation 6cos²x + cosx − 2 = 0 for 0° ≤ x ≤ 360°. Give answers to 1 decimal place.

Let t = cosx: 6t² + t − 2 = 0 → (2t − 1)(3t + 2) = 0 [M1 factorise]
cosx = 1/2 → x = 60°, 300° [A1]
cosx = −2/3: α = arccos(2/3) ≈ 48.2°. cos negative → 2nd and 3rd quadrant [M1]
x = 180° − 48.2° = 131.8° and x = 180° + 48.2° = 228.2° [A1 A1]
All four values correct [A1]
Answer: x = 60°, 131.8°, 228.2°, 300°

Past Paper Q5 — 9709/12 style [7 marks]

(a) Prove that (sinθ + cosθ)² = 1 + 2sinθcosθ. [2 marks]
(b) Solve (sinθ + cosθ)² = 1 + sinθ for 0° ≤ θ ≤ 360°. [5 marks]

(a) (sinθ + cosθ)² = sin²θ + 2sinθcosθ + cos²θ = 1 + 2sinθcosθ [M1 expand, A1 use identity]
(b) From (a): 1 + 2sinθcosθ = 1 + sinθ → 2sinθcosθ = sinθ [M1]
2sinθcosθ − sinθ = 0 → sinθ(2cosθ − 1) = 0 [M1 factorise]
sinθ = 0 → θ = 0°, 180°, 360° [A1]
cosθ = 1/2 → θ = 60°, 300° [A1]
All five values stated [A1]
Answer: θ = 0°, 60°, 180°, 300°, 360°

Solving Trigonometric Equations A-Level Pure 1