Further trig equations combine every identity you know: sec² and cosec² identities, double angle formulae, the addition formulae and the R form. The skill is diagnosis. Read the equation, decide which identity reduces it to a single trig function of a single angle, then solve and check you have found every solution in the interval.
The toolkit for harder trig equations
- sin²θ + cos²θ = 1
- sec²θ = 1 + tan²θ and cosec²θ = 1 + cot²θ
- sin 2A = 2 sin A cos A; cos 2A = 1 − 2 sin²A = 2 cos²A − 1; tan 2A = 2 tan A / (1 − tan²A)
- cot x + tan x = 2 cosec 2x and cot x − tan x = 2 cot 2x
- a cos θ + b sin θ = R cos(θ − α)
How do you choose which identity to use?
Before writing anything, ask what the equation is made of. That tells you the route:
- sec² with tan, or cosec² with cot: use the Pythagorean identity to get a quadratic
- cos 2x with a stray sin x or cos x: use the double angle form that matches
- sin 2x with sin x or cos x: expand, then factorise rather than divide
- sin(x + α) mixed with sin x: expand with the addition formula
- a cos x + b sin x = c: use the R form
- every term of degree 2 (sin²x, sin x cos x, cos²x): divide by cos²x to get a quadratic in tan x
The aim is always one trig function of one angle. If your first step leaves two different trig functions in the equation, stop and re-diagnose rather than pressing on.
Equations with sec, cosec, cot and double angles
The identities sec²(kx) = 1 + tan²(kx) and cosec²(kx) = 1 + cot²(kx) work for any angle, including 2x. Substitute u = 2x, transform the interval, and the equation becomes an ordinary quadratic. Then flip each root back: cot u = k means tan u = 1/k.
The tan double angle formula can turn an equation into a cubic in tan x. A cubic can have fewer real roots than you expect, so check each factor.
When reciprocal expressions become double angles
Combining cot x + tan x over a common denominator gives (cos²x + sin²x) / (sin x cos x) = 1 / (sin x cos x). Since sin x cos x = ½ sin 2x, this is 2 / sin 2x, or 2 cosec 2x. Similarly, cot x − tan x = 2 cot 2x.
When you multiply an equation by sin x or cos x to clear a fraction, you can create roots that fail the original equation, because the original was undefined there. The function sec x is undefined when cos x = 0, while cosec x and cot x are undefined when sin x = 0. Reject any root that lands on those.
How do you make sure you have every solution?
Most lost marks in this topic are missing solutions, not wrong ones. Four habits stop that:
- Factorise, never divide by a trig function.
- Transform the interval when the angle is 2x or x + α, because a doubled angle doubles the number of solutions.
- Keep the ± when you square root.
- Check for impossible roots (such as sin x = 2) and undefined ones.
When you finish, count your answers and ask whether that number makes sense for the structure of the equation.
Worked examples
Example 1. Solve sec²x = 4 tan x for 0° ≤ x < 360°.
- Diagnose: sec² with tan, so use sec²x = 1 + tan²x.
- Substitute and collect: tan²x − 4 tan x + 1 = 0.
- Quadratic formula: tan x = (4 ± √12) / 2 = 2 ± √3.
- Recognise exact values: 2 + √3 = tan 75° and 2 − √3 = tan 15°.
- Tangent repeats every 180°: x = 75°, 255° and x = 15°, 195°.
Answer: x = 15°, 75°, 195°, 255°.
Marks: M1 for the identity, M1 for solving the quadratic, A1 for all four.
Example 2. Solve cosec²x = 2 cot x + 4 for 0° ≤ x < 360°, to 1 d.p. where appropriate.
- Use cosec²x = 1 + cot²x: 1 + cot²x = 2 cot x + 4.
- Collect: cot²x − 2 cot x − 3 = 0.
- Factorise: (cot x − 3)(cot x + 1) = 0, so cot x = 3 or cot x = −1.
- Flip each root: tan x = 1/3 or tan x = −1.
- tan x = 1/3 gives x = 18.4°, 198.4°. tan x = −1 gives x = 135°, 315°.
Answer: x = 18.4°, 135°, 198.4°, 315°.
Marks: M1 for the identity, M1 for factorising, M1 for flipping to tan, A1 for all four.
Example 3. Solve cot x + tan x = 4 for 0° ≤ x < 360°.
- Combine: cos x / sin x + sin x / cos x = 1 / (sin x cos x).
- Since sin x cos x = ½ sin 2x, the left side is 2 / sin 2x.
- So 2 / sin 2x = 4, giving sin 2x = ½.
- Let u = 2x, with 0° ≤ u < 720°: u = 30°, 150°, 390°, 510°.
- Halve: x = 15°, 75°, 195°, 255°.
Answer: x = 15°, 75°, 195°, 255°.
Marks: M1 for combining, M1 for spotting the double angle, A1 for all four.
Example 4. Solve sin 2x = tan x for 0° ≤ x < 360°.
- Write tan x as a fraction: 2 sin x cos x = sin x / cos x.
- Multiply by cos x: 2 sin x cos²x = sin x.
- Factorise: sin x(2 cos²x − 1) = 0.
- sin x = 0 gives x = 0°, 180°.
- 2 cos²x − 1 = cos 2x, so cos 2x = 0. With 0° ≤ 2x < 720°: 2x = 90°, 270°, 450°, 630°, so x = 45°, 135°, 225°, 315°.
Answer: Six solutions: x = 0°, 45°, 135°, 180°, 225°, 315°.
Marks: M1 for clearing the fraction, M1 for factorising, A1 for both branches, A1 for all six.
Common mistakes
- Dividing by a trig function. Dividing by sin x loses every solution where sin x = 0. Bring everything to one side and factorise.
- Not transforming the interval. If the angle is 2x on 0° ≤ x < 360°, solve for 2x on 0° ≤ 2x < 720°, then halve.
- Forgetting to flip cot back to tan. cot x = 3 means tan x = 1/3, not tan x = 3.
- Keeping roots where the original equation is undefined. After multiplying through by sin x or cos x, check each answer in the original equation.
- Choosing an identity that leaves two trig functions. If you still have both sin x and cos x after your first step, go back and pick a different identity.
How harder trig equations are examined
In Edexcel International A-Level these questions appear throughout Pure 3 (paper WMA13), mostly in the chapters on trigonometric functions and addition formulae. In Edexcel A-Level (9MA0) they are Year 2 Pure. Expect:
- a proof in part (a) followed by “hence solve” in part (b)
- equations mixing sec, cosec or cot with tan or cot
- double angle equations where the interval must be transformed
- questions that ask for all solutions, with an impossible or undefined root to reject
For timed practice with mark schemes, use the Pure 3 trigonometric addition formulae exam questions and the Pure 3 trigonometric functions exam questions. The A-Level trigonometry revision engine generates unlimited extra equations.
If a technique feels shaky, revisit sec, cosec and cot, the double angle formulae or the R formula.
Frequently asked questions
How do I know which trig identity to use?
Look at what the equation contains. sec² with tan points to sec² = 1 + tan²; cos 2x with sin x points to cos 2x = 1 − 2 sin²x; a cos x + b sin x points to the R form. Pick the identity that leaves one trig function of one angle.
Why do I keep missing solutions?
Usually because of dividing by a trig function, not transforming the interval for 2x, or dropping the ± after a square root. Count your answers at the end and check the number makes sense.
When do I have to reject a solution?
Reject a root that is impossible, such as sin x = 2, or one where the original equation is undefined, such as sec x when cos x = 0. Always say why in the exam.