Solve quadratic trigonometric equations and reject invalid ratios
Use an identity to form a quadratic in one ratio, check that its roots are possible and recover every permitted angle.
Pure Mathematics 1 (Paper 1) · AS & A Level Mathematics 9709, 2026–2027. State angle units and find every solution in the supplied interval; general forms of solution are not required in section 1.5. Written practice is self-checked.
Before you begin
- Solve quadratic equations and use sin² x + cos² x = 1.
- Find all simple trigonometric solutions in a supplied interval.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
The substitution u = cos x gives u = 3/2 or u = −1/2. Which roots can give real angles?
What you will learn
- Turn an equation in trigonometric ratios into a quadratic in one variable.
- Reject impossible ratio values and find all remaining angles in the stated interval.
A quadratic root is a ratio before it is an angle
For 2 sin² x − 3 sin x + 1 = 0, let u = sin x and solve 2u² − 3u + 1 = (2u − 1)(u − 1) = 0. The ratio values are u = 1/2 or 1, both allowed because sine lies in [−1, 1]. On 0° ≤ x ≤ 360°, they give x = 30°, 150° and 90° respectively. List distinct angles once and check them in the original equation.
Dividing by sin x or cos x can remove legitimate zero-ratio solutions. For example sin x(2 sin x − 1) = 0 requires both sin x = 0 and sin x = 1/2. On 0° ≤ x ≤ 360°, its roots are 0°, 30°, 150°, 180° and 360°. Factor first and examine each factor, rather than dividing by a quantity that might be zero.
Use an identity when two ratios occur
For 3 sin² x + 2 cos x − 1 = 0, replace sin² x by 1 − cos² x and let u = cos x. The equation becomes 3u² − 2u − 2 = 0, giving u = (1 ± √7)/3. The plus root is greater than 1 and is impossible for a real cosine. The minus root is about −0.548584 and produces two angles in a full turn. Preserve its exact value until the final inverse calculation.
For tan² x − 3 tan x + 2 = 0, factor to obtain tangent values 1 and 2. Tangent is not bounded by [−1, 1], so 2 is valid; instead check its excluded inputs and the requested interval. On 0° ≤ x < 180°, the answers are 45° and approximately 63.4°. A range restriction belongs to a particular ratio, rather than every trigonometric function.
Pause and explain
Why is dividing sin x(2 sin x − 1) = 0 by sin x unsafe?
Worked example
Solve 2 sin² x − 3 sin x + 1 = 0 for 0° ≤ x ≤ 360°.
Show the worked solution
- Let u = sin x, then factor (2u − 1)(u − 1) = 0.
- Both u = 1/2 and u = 1 lie in the sine range; the first gives 30° and 150°, and the second gives 90°.
- Check all three in the original equation and list them in increasing order without duplicates.
Answer x = 30°, 90° or 150°.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
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 2 sin² x − 3 sin x + 1 = 0 on 0° ≤ x ≤ 360°.
Give me a hint
Solve a quadratic in u = sin x, then recover all angles for each allowed ratio.
Compare my reasoning
- Factor (2u − 1)(u − 1) = 0, giving u = 1/2 or 1.
- Sine 1/2 gives 30° and 150°; sine 1 gives 90° within this interval.
- Substitute all three and list x = 30°, 90°, 150°; neither endpoint satisfies the equation.
x = 30°, 90° or 150°.
Look for these in your work
- Find both allowed quadratic ratio roots.
- Recover and check all angles for each root.
Solve 2 cos² x − 5 cos x + 2 = 0 for 0 ≤ x < 2π.
Give me a hint
Let u = cos x, factor the quadratic and apply the cosine range before taking inverses.
Compare my reasoning
- Factor (2u − 1)(u − 2) = 0, giving u = 1/2 or 2.
- Reject u = 2 because a real cosine cannot exceed 1.
- The remaining equation cos x = 1/2 gives x = π/3 and 5π/3 in the specified interval.
x = π/3 or 5π/3; the cosine root 2 is rejected.
Look for these in your work
- Separate algebraic roots from permitted cosine values.
- Find both full-turn angles for the valid root.
Solve sin² x + cos x = 1 for 0° ≤ x ≤ 360°. Explain why dividing by cos x would lose answers.
Give me a hint
Replace sin² x with 1 − cos² x and factor without division.
Compare my reasoning
- Substitute the identity to get 1 − cos² x + cos x = 1, hence cos x(1 − cos x) = 0.
- For cos x = 0 the angles are 90° and 270°; for cos x = 1 they are the two included endpoints 0° and 360°.
- All four satisfy the original equation. Dividing by cos x would discard 90° and 270° because their cosine is zero.
x = 0°, 90°, 270° or 360°; division by cosine loses the two zero-cosine answers.
Look for these in your work
- Use the squared identity and preserve both factors.
- Include valid endpoints and explain the lost-root danger.
Common mistakes
- Quadratic roots are ratio values; convert them into all permitted angles.
- Do not divide by a ratio that could be zero and lose its solutions.
- Reject sine or cosine values outside [−1, 1], but do not apply that restriction to tangent.
What can you explain now?
Solve sin x(2 sin x − 1) = 0 for 0° ≤ x ≤ 360°.
Compare with the explanation
x = 0°, 30°, 150°, 180° or 360°.
The zero factor supplies the three zeros, while the second factor supplies both angles with sine 1/2.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Give a complete routine for a quadratic trigonometric equation: substitute, solve, range-check, recover angles, check domains and endpoints, then verify in the original equation.
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.