Solve sine, cosine and tangent equations on a domain
Find every solution to basic sine, cosine and tangent equations in a stated interval using exact angles or a calculator reference angle.
Cambridge IGCSE Additional Mathematics 0606, 2028–2030 syllabus, Version 1. Paper 1 is non-calculator; Paper 2 requires a scientific calculator. These original lessons provide self-checked topic practice, not a Cambridge grade or a timed paper. Full-paper assessment and human specialist review are still outstanding.
Before you begin
- Cambridge assumes knowledge of IGCSE Mathematics or an equivalent course. Check the domain before using an inverse or logarithm, and retain exact values where the question asks for them.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
In which two quadrants is sine positive?
What you will learn
- Solve equations involving sine, cosine or tangent on a specified domain
- Use the unit circle to include every solution in the interval
- State exact or appropriately rounded angles and exclude undefined values
Solve for the function value before finding angles
Rearrange the equation to isolate sin θ, cos θ or tan θ. Use a reference angle, the sign pattern in each quadrant and the stated interval to list every angle.
For sine, the same positive value occurs in Quadrants I and II, and a negative value occurs in III and IV. For cosine, matching values occur in I/IV or II/III; for tangent, each value repeats every 180° (π radians).
Use the interval to control the answer
A calculator's inverse function usually returns one principal angle. Use the quadrant pattern or period to find the other solutions in the requested domain, then remove any that fall outside it.
Check open versus closed endpoints. If a coefficient or substitution changes the angle, transform the domain too. Give exact special angles where possible; otherwise use the requested rounding and include the unit.
When a denominator contains a trigonometric function, values that make it zero are excluded even if an algebraic rearrangement appears to allow them.
Pause and explain
What is the reference angle for tan θ = −1?
Worked example
Solve 2 sin θ = √3 for 0 ≤ θ < 2π.
Show the worked solution
- Divide by 2 to get sin θ = √3/2. The reference angle is π/3.
- Sine is positive in Quadrants I and II, so the angles in one full turn are π/3 and π − π/3.
- Both lie in the stated half-open interval.
Answer θ = π/3 or 2π/3.
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 sin θ = −1/2 for 0 ≤ θ < 2π.
Give me a hint
Sine is negative in Quadrants III and IV; the reference angle is π/6.
Compare my reasoning
- The reference angle for 1/2 is π/6.
- The Quadrant III solution is π + π/6 = 7π/6.
- The Quadrant IV solution is 2π − π/6 = 11π/6.
θ = 7π/6 or 11π/6.
Look for these in your work
- Reference angle correct
- Both negative-sine quadrants used
- Solutions lie in the interval
Solve 2 cos θ + 1 = 0 for 0 ≤ θ ≤ 2π.
Give me a hint
Rearrange to cos θ = −1/2 and include the closed endpoint only if it satisfies the equation.
Compare my reasoning
- cos θ = −1/2 has reference angle π/3.
- Cosine is negative in Quadrants II and III.
- The angles are 2π/3 and 4π/3; neither endpoint adds another solution.
θ = 2π/3 or 4π/3.
Look for these in your work
- Equation isolated correctly
- Quadrants II and III selected
- Endpoints checked
Solve 5 sin 3θ + 2 cos 3θ = 0 for 0 ≤ θ < 2π.
Give me a hint
First show cos 3θ cannot be zero at a solution; then solve for tan 3θ on 0 ≤ 3θ < 6π.
Compare my reasoning
- If cos 3θ = 0, then sin 3θ = ±1 and the left side is ±5, so no solution is lost by dividing by cos 3θ.
- Thus tan 3θ = −2/5. Let α = arctan(2/5); in 0 ≤ 3θ < 6π the solutions are 3θ = π − α + kπ for k = 0, 1, 2, 3, 4, 5.
- Divide by 3; each resulting θ lies in the requested half-open interval.
θ = (π − α + kπ)/3 for k = 0, 1, 2, 3, 4, 5, where α = arctan(2/5).
Look for these in your work
- Possible zero denominator checked
- All six-periodic angle solutions listed
- Transformed domain and interval checked
Common mistakes
- Giving only the principal calculator angle
- Using the wrong quadrant pattern for the function
- Changing between degrees and radians part-way through
- Including a value outside an open endpoint or where a denominator is zero
What can you explain now?
How many solutions does cos θ = −1/2 have on 0 ≤ θ < 2π?
Compare with the explanation
Two
Cosine is negative in Quadrants II and III, giving 2π/3 and 4π/3.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
How do a reference angle, quadrant signs, periodicity and the stated interval combine to give every solution?
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.