Trigonometric equations
Find every solution in an interval by considering signs and symmetry.
Pure Mathematics 1 (Paper 1) · Mathematics 9709, 2026–2027. This revision lesson focuses on complete solution sets in specified intervals, not general forms of solution or the whole trigonometry syllabus. Written solutions use self-check criteria, not an official mark scheme. Specialist accuracy review has not been completed.
Before you begin
- Use exact sine and cosine values for 30°, 60° and their radian equivalents.
- Factor a product equal to zero without dividing by a possibly zero factor.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Solve sin θ = −√3/2 for 0° ≤ θ < 360°. Which set includes every solution?
What you will learn
- Find a reference angle.
- Select all quadrants consistent with the equation.
Isolate the trigonometric ratio
Rearrange the equation until a single sine, cosine or tangent is isolated. Find a reference angle using an exact value or an inverse function. An inverse calculator result gives one value, not automatically every solution in the requested interval.
Sine is positive in quadrants I and II, cosine in I and IV, and tangent in I and III. Use the sign together with the reference angle, then check the original equation.
Respect the interval and units
A solution outside the stated interval must be excluded. Endpoints need particular care because an inclusive interval may contain both 0° and 360°. Check whether the question uses degrees or radians before forming solutions.
Sources for this lesson
Pause and explain
The cosine graph has value 1 at 0° and 360°. For cos x = 1 on 0 ≤ x ≤ 2π radians, which solution set is complete?
Worked example
Solve 2cos θ=−1 for 0°≤θ≤360°.
Show the worked solution
- Rearrange to cos θ=−1/2, with reference angle 60°.
- Cosine is negative in quadrants II and III.
- The angles are 180°−60° and 180°+60°.
Answer 120° and 240°
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 sin θ + √3 = 0 for −π ≤ θ ≤ π. State exact radian answers and explain why a positive reference angle is not automatically a solution.
Give me a hint
Isolate sine. Locate the two negative-sine positions within the supplied interval; use a full-turn shift if starting with positive angles.
Compare my reasoning
- Rearrange to sin θ = −√3/2. Its reference angle is π/3, but sin(π/3) is positive and does not satisfy this equation.
- In one positive turn the corresponding angles are 4π/3 and 5π/3. Subtract 2π from each to place them in the requested interval: −2π/3 and −π/3.
- Both have sine −√3/2. At −π, 0 and π the sine is 0, so these points are not solutions. The negative half-turn contains exactly the two required crossings.
θ = −2π/3 or −π/3.
Look for these in your work
- Isolate the ratio and keep its negative sign.
- Convert all selected angles to the supplied interval and units.
- Check both original substitution and completeness.
Solve cos(2x) = −√3/2 for 0 ≤ x < 2π. Show the interval for u = 2x before finding x.
Give me a hint
The cosine input covers two full turns. Find every permitted u first, then divide all values by 2.
Compare my reasoning
- Let u = 2x. Then 0 ≤ u < 4π, so searching only 0 ≤ u < 2π would miss a full turn.
- In the first turn, negative cosine with reference angle π/6 gives u = 5π/6 and 7π/6. In the second turn add 2π to obtain 17π/6 and 19π/6.
- Divide each angle by 2: x = 5π/12, 7π/12, 17π/12, 19π/12. All lie in the original half-open interval and each doubled angle gives the required cosine.
x = 5π/12, 7π/12, 17π/12 or 19π/12.
Look for these in your work
- Transform the interval with the argument.
- Include both negative-cosine positions in both turns.
- Divide every angle and filter in the original interval.
Solve sin x(2 cos x + 1) = 0 for −π ≤ x ≤ π. A learner divides by sin x before solving. Identify exactly which solutions this loses.
Give me a hint
Use the zero-product rule and solve both factors separately; retain included endpoints even when they share a unit-circle direction.
Compare my reasoning
- The product is zero when sin x = 0 or 2 cos x + 1 = 0. Dividing by sin x would exclude the entire first case.
- For sin x = 0, the allowed values are −π, 0 and π. For cos x = −1/2, the allowed values are −2π/3 and 2π/3.
- Combine and order the values: −π, −2π/3, 0, 2π/3, π. Each makes at least one original factor zero. Division by sin x loses −π, 0 and π.
x = −π, −2π/3, 0, 2π/3 or π. Division loses −π, 0 and π.
Look for these in your work
- Preserve both branches of the zero-product rule.
- Find all values in each branch, including endpoints.
- Name the lost values and check them in the original product.
Common mistakes
- One inverse-trig output is not a complete solution set.
- Do not mix radian and degree values.
What can you explain now?
Solve sin θ=1/2 for 0≤θ≤2π.
Compare with the explanation
π/6 and 5π/6
The positive sine solutions lie in quadrants I and II.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Explain why an inverse-function output may be only one solution. Describe how to transform the interval when the input is 2x, and why dividing by sin x needs a separate zero-sine case. Use a sketch to distinguish the endpoints 0 and 2π from the half-turn π.
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.