Solving trigonometric equations
Because trig functions are periodic, an equation has infinitely many solutions and the question always restricts to an interval. The inverse function gives just ONE solution — the principal value — and you must generate the rest using the symmetry of the graph and the period. Finding only the calculator answer is the classic way to lose most of the marks.
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Method: how to approach it
The order below is what examiners expect to see, and each step carries its own marks.
- Isolate the trig functionGet to sin x = k, cos x = k or tan x = k before doing anything else.
- Find the principal valueUse arcsin, arccos or arctan for one solution.
- Generate the second solution in the cycleFor sine, the partner is 180° − x; for cosine, 360° − x (i.e. −x); tangent repeats every 180° so it has one per period.
- Add multiples of the period and filterAdd 360° (or the adjusted period for sin(bx)) repeatedly, keeping every solution inside the stated interval.
Worked example
Solve 2 sin x = 1 for 0° ≤ x ≤ 360°.
- Isolate: sin x = 1/2.
- Principal value: x = arcsin(1/2) = 30°.
- Sine is also positive in the second quadrant: x = 180° − 30° = 150°.
- Adding 360° would exceed the interval.
Answer. x = 30° or x = 150°.
Where marks get dropped
These are the specific errors that cost credit on solving trigonometric equations questions — QED's rubric penalises each of them separately.
- Giving only the calculator’s principal value. Every interval question expects the complete solution set.
- Forgetting to adjust the interval for a multiple angle. For sin 2x = k with 0 ≤ x ≤ 360°, solve for 2x over 0 to 720° first, then halve.
- Dividing by a trig function and losing solutions. From sin x cos x = sin x, factor instead: sin x(cos x − 1) = 0.
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Solving trigonometric equations — frequently asked questions
How many solutions should I expect?
Roughly two per period for sine and cosine, one per period for tangent. For sin(bx) over a full turn, expect about 2b solutions.
How do I handle sin 2x = 1/2 on 0 ≤ x ≤ 360°?
Let u = 2x, solve over 0 ≤ u ≤ 720°, getting u = 30, 150, 390, 510, then halve each: x = 15°, 75°, 195°, 255°.
What if the equation is quadratic in sin x?
Substitute s = sin x, solve the quadratic, then solve each resulting sin x = value separately — discarding any root outside [−1, 1].
The rest of Trigonometry
Triangles, the unit circle, identities and periodic functions. Each subtopic below has its own method, worked example and mark-losing traps.
- 1Right-triangle ratios: sin, cos & tan
- 2The unit circle & exact values
- 3Radians, degrees & arc length
- 4Graphs of sin, cos & tan
- 5Pythagorean & reciprocal identities
- 6Angle-sum, difference & double-angle formulas
- 7Solving trigonometric equations
- 8The sine rule & the cosine rule
- 9Triangle area, sectors & segments
- 10Inverse trigonometric functions
- 11Proving trigonometric identities
- 12Polar coordinates & converting to Cartesian
- 13Modelling periodic phenomena with sinusoids
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