Solve phase shifted trigonometric equations over a full interval
Turn a sine-cosine equation into one shifted ratio, use its range and retain every branch inside the original interval.
Section 3.3 · Trigonometry
Exam scope and assessment notes
Pure Mathematics 2 (Paper 2, AS only) and Pure Mathematics 3 (Paper 3, A Level) · Mathematics 9709, 2026–2027. Paper 1 knowledge is assumed. The AS route using Papers 1 + 2 cannot be carried forward to A Level. Shared sections 2.3 and 3.3 cover all six trigonometric functions, reciprocal squared identities, compound and double angles, and amplitude-phase forms used in simplification, exact evaluation and equations. State angle units, keep original domains and find every solution in the supplied interval. Written practice is self-checked.
Before you begin
- Match amplitude-phase coefficients and use principal inverse trigonometric values.
- Find every equation solution in a shifted interval without rounding the phase prematurely.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Why does 3 sin θ + 4 cos θ = 6 have no real solution?
What you will learn
- Use amplitude-phase forms to solve mixed sine and cosine equations over a supplied interval.
- Distinguish impossible targets, a single extremum solution and two ordinary full-turn solutions.
Check the target range before using an inverse key
An equation a sin θ + b cos θ = c can be rewritten using a positive amplitude R = √(a² + b²). If |c| > R there is no real solution. If |c| = R, the target is an extremum and the two usual inverse branches merge into one distinct angle per full turn. For |c| < R, a non-zero amplitude gives two distinct angles per full turn; both must still be filtered by any shorter interval. The case R = 0 is the constant expression and must be checked directly.
For 3 cos θ + 2 sin θ = 1, use √13 cos(θ − β) = 1, with β = arctan(2/3) acute. Let γ = cos⁻¹(1/√13). The cosine equation has shifted-angle values ±γ with whole turns added as needed. On 0° ≤ θ < 360°, the resulting angles are β + γ ≈ 107.6° and 360° + β − γ ≈ 319.8°. Keep β and γ unrounded until the final one-decimal answers, then verify using their exact values in the original expression.
Move the interval together with the angle
For 3 sin θ + 4 cos θ = −4, use 5 sin(θ + α) = −4 with α = arctan(4/3) ≈ 53.1°. The shifted input u = θ + α runs through [α, 360° + α). In this interval the sine −4/5 values are u = 180° + α and 360° − α. Subtract α to get θ = 180° and 360° − 2α ≈ 253.7°. A principal inverse-sine result alone would not supply both branches or correctly shifted interval values.
For a supplied displacement d(θ) = 5 + 3 sin θ + 4 cos θ cm on 0 ≤ θ < 2π radians, setting d = 8 gives sin(θ + α) = 3/5, now with α = arctan(4/3) in radians. Write β = sin⁻¹(3/5) = arctan(3/4). The value u = β lies below the shifted interval's lower bound α, so use u = 2π + β as well as u = π − β. Subtracting α gives θ = π/2 and θ = 2π + β − α ≈ 6.00 radians. The whole-turn adjustment recovers a valid solution near the end of the original interval.
Pause and explain
How many distinct angles solve 3 sin θ + 4 cos θ = 5 in 0° ≤ θ < 360°?
Worked example
Solve 3 cos θ + 2 sin θ = 1 for 0° ≤ θ < 360°, giving final angles to one decimal place.
Show the worked solution
- Rewrite as √13 cos(θ − β) = 1, with β = arctan(2/3), and let γ = cos⁻¹(1/√13).
- Use both cosine branches and add a full turn to the negative recovered angle: θ = β + γ or 360° + β − γ.
- The values are approximately 107.6° and 319.8°; use the unrounded values to check the original equation and half-open interval.
Answer θ = β + γ or 360° + β − γ, where β = arctan(2/3), γ = cos⁻¹(1/√13); approximately 107.6° or 319.8°.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Solve 3 sin θ + 4 cos θ = 5 for 0° ≤ θ < 360°. State why there is only one distinct solution.
Give me a hint
Use 5 cos(θ − β) with β = arctan(3/4); the target is the maximum value.
Compare my reasoning
- The equation becomes cos(θ − β) = 1.
- The only recovered angle in the half-open full turn is θ = β = arctan(3/4) ≈ 36.9°.
- Both usual inverse branches give the same maximum angle, so it is counted once; the exact value gives 3 sin θ + 4 cos θ = 5.
θ = arctan(3/4) ≈ 36.9°; one distinct maximum solution.
Look for these in your work
- Identify an attained amplitude endpoint before solving.
- Avoid duplicating a merged inverse branch.
Solve 3 sin θ + 4 cos θ = −4 for 0° ≤ θ < 360°, to one decimal place for non-exact angles.
Give me a hint
Use α = arctan(4/3) and solve sin(θ + α) = −4/5 over the shifted interval.
Compare my reasoning
- The shifted interval is α ≤ u < 360° + α, and its two sine solutions are u = 180° + α and 360° − α.
- Subtract α to get θ = 180° or 360° − 2α.
- The second angle is approximately 253.7°; exact substitution verifies both and neither lies at the excluded 360° endpoint.
θ = 180° or 360° − 2 arctan(4/3) ≈ 253.7°.
Look for these in your work
- Keep both shifted sine branches in the correct interval.
- Subtract the exact phase and verify in the original equation.
A supplied displacement is d(θ) = 5 + 3 sin θ + 4 cos θ cm for 0 ≤ θ < 2π radians. Find all angles where d = 8 cm, using α = arctan(4/3) and β = sin⁻¹(3/5).
Give me a hint
Solve sin(θ + α) = 3/5; its smaller principal angle falls below the shifted interval, so consider one added turn.
Compare my reasoning
- The shifted input u lies in [α, 2π + α), and the two permitted values are π − β and 2π + β.
- Subtract α to obtain θ = π − β − α = π/2, since α + β = π/2, and θ = 2π + β − α.
- The second value is approximately 6.00 radians; both exact values give d = 8 and belong to the original interval.
θ = π/2 or 2π + β − α ≈ 6.00 radians, with α = arctan(4/3), β = sin⁻¹(3/5).
Look for these in your work
- Adjust a shifted solution by a whole turn when the principal value falls outside the interval.
- Check the displacement, radian units and original angle bounds.
Common mistakes
- An inverse key returns one principal value, not the full interval solution set.
- Shift the interval with the input and include whole turns where needed.
- Do not round a phase before calculating the final angles, or count the same extremum angle twice.
What can you explain now?
Can 3 sin θ + 4 cos θ = 6 have a real solution? Explain without using an inverse key.
Compare with the explanation
No; the expression has range [−5, 5].
Its amplitude is 5, so a target value 6 lies outside the possible range.
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Make it stick
Use the amplitude to distinguish no solution, one maximum solution and two ordinary solutions. Explain how to find both branches when a shifted interval crosses the end of a turn.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.