Find every trigonometric solution in a stated interval
Use reference angles and periodic graphs, enlarge the interval for a multiplied argument and check endpoints before listing solutions.
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
- Use principal inverse values and trigonometric graph periods.
- Solve inequalities for a transformed angle and interpret open or closed endpoints.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
If 0 ≤ x ≤ π and u = 2x, what is the interval for u?
What you will learn
- Find all simple sine, cosine and tangent equation solutions within a supplied interval.
- Transform the full argument interval before solving an equation containing a multiplied angle.
Use the interval and the graph to complete the answer
For sin x = 1/2 on 0° ≤ x ≤ 360°, the reference angle is 30°. Sine is positive in quadrants I and II, so x = 30° or 150°. For cosine use its horizontal-coordinate sign, and for tangent use the sine/cosine quotient sign with period 180°. For tan x = −1 on −180° < x ≤ 180°, the complete answer is −45° or 135°.
Check interval endpoints explicitly. sin x = 0 on 0° ≤ x ≤ 360° has solutions 0°, 180° and 360°; replacing the final ≤ by < removes 360°. Do not combine equivalent directions if they are distinct numbers in the requested interval. Values such as sin x = 1.1 have no real solutions; an inverse-key error is then evidence of an impossible ratio.
Solve in the argument's full interval, then undo the change
For cos 2x = 1/2 with 0 ≤ x ≤ π, put u = 2x, so 0 ≤ u ≤ 2π. Within this enlarged interval, u = π/3 or 5π/3. Divide both values by 2 to obtain x = π/6 or 5π/6. Solving only for one principal u loses a valid answer; solving over the original x interval as if it belonged to u can also lose answers.
For sin 2x = −1/2 with 0° ≤ x < 360°, the argument covers 0° ≤ u < 720°, so it contains two full sine periods. List the four negative-sine arguments 210°, 330°, 570° and 690° before halving them. If there is an input shift, transform both interval endpoints too. This section requests solutions in specified intervals, so a general infinite solution formula is unnecessary.
Pause and explain
Solve sin x = 0 on 0° ≤ x ≤ 360°.
Worked example
Solve cos 2x = 1/2 for 0 ≤ x ≤ π.
Show the worked solution
- Let u = 2x and transform the interval to 0 ≤ u ≤ 2π.
- Cosine is positive at the π/3 reference angle in quadrants I and IV, so u = π/3 or 5π/3.
- Divide each by 2, check the original interval and substitute to verify both.
Answer x = π/6 or 5π/6.
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 cos x = −√2/2 for 0 ≤ x < 2π.
Give me a hint
Use a π/4 reference angle and the quadrants with negative cosine.
Compare my reasoning
- The exact magnitude √2/2 gives reference angle π/4.
- Negative cosine occurs in quadrants II and III, giving x = π − π/4 = 3π/4 and x = π + π/4 = 5π/4.
- Both values are in the interval and give the required ratio; neither endpoint is a solution.
x = 3π/4 or 5π/4.
Look for these in your work
- Choose both quadrants for the required sign.
- Check the interval and original ratio.
Solve sin 2x = −1/2 for 0° ≤ x < 360°.
Give me a hint
Use u = 2x and find all negative-sine values over two full turns.
Compare my reasoning
- The argument interval is 0° ≤ u < 720°.
- The solutions for u are 210°, 330°, 570° and 690°.
- Halve every value to obtain x = 105°, 165°, 285° and 345°; all satisfy the original interval.
x = 105°, 165°, 285° or 345°.
Look for these in your work
- Expand the argument interval before using periodicity.
- Retain every solution when dividing back to x.
A supplied model is h(t) = 2 + 3 sin(πt/6) metres for 0 ≤ t ≤ 12 seconds. Find every time when h = 3.5 metres.
Give me a hint
Isolate the sine value, then transform the time interval into an angle interval.
Compare my reasoning
- Set 2 + 3 sin(πt/6) = 3.5 to obtain sin(πt/6) = 1/2.
- For u = πt/6, the time interval gives 0 ≤ u ≤ 2π, with solutions u = π/6 and 5π/6.
- Use t = 6u/π to get 1 and 5 seconds; substitute each in the original height model.
t = 1 s or 5 s.
Look for these in your work
- Isolate the ratio without confusing the height with the sine output.
- Convert all permitted angles back into times with units.
Common mistakes
- The principal inverse value is only a starting angle.
- Transform the whole interval when the argument is multiplied or shifted.
- Check open endpoints, closed endpoints and excluded tangent inputs.
What can you explain now?
Solve sin x = 0 for 0° ≤ x ≤ 360°.
Compare with the explanation
x = 0°, 180° or 360°.
The graph has zeros at both included endpoints as well as at the half turn.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Explain why sin 2x can have four answers over one full turn of x. Describe the interval change and endpoint check before any inverse calculation.
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.