IGCSE Additional Mathematics 0606 · Topic 11

IGCSE Additional Mathematics: Trigonometry Practice Questions

Trigonometric equations have infinitely many solutions, so a range is always specified. Find the first solution with the inverse function, then use the symmetry of the graph to find every other solution in that range.

Cambridge IGCSE Additional Mathematics (0606) · Topic 11: Trigonometry

Topic 11 of Cambridge IGCSE Additional Mathematics 0606 is where candidates most often lose marks by giving only one solution. Sketching the relevant curve across the stated range makes the full set obvious. The questions below train that habit.

What you need to know for Trigonometry

IGCSE Additional Mathematics Trigonometry questions and answers

4 exam-style questions written to the 0606 syllabus. Try each one on paper first, then open the worked answer to check your method against the marks.

Question 1[4 marks]

Solve 2 sin x = 1 for 0 degrees less than or equal to x, less than or equal to 360 degrees.

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Answer: x = 30 degrees or x = 150 degrees
  1. Rearrange: sin x = 0.5.
  2. The first solution is x = the inverse sine of 0.5 = 30 degrees.
  3. Sine is also positive in the second quadrant, where the solution is 180 minus 30 = 150 degrees.
  4. No further solutions lie in the range, since sine is negative between 180 and 360 degrees.
How the marks are awarded. 1 mark for sin x = 0.5. 1 mark for x = 30 degrees. 1 mark for using the symmetry of the sine curve. 1 mark for x = 150 degrees.
Where students lose the mark. Giving only 30 degrees. The range covers a full revolution, so every solution within it must be listed.
Question 2[5 marks]

Solve 3 tan2 x minus 1 = 0 for 0 degrees less than or equal to x, less than or equal to 360 degrees.

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Answer: x = 30, 150, 210 and 330 degrees
  1. Rearrange: tan2 x = one third, so tan x = plus or minus 1 over the square root of 3.
  2. Take the positive case first: tan x = 0.5774, giving x = 30 degrees. Tangent repeats every 180 degrees, so also x = 210 degrees.
  3. Now the negative case: tan x = minus 0.5774, giving a first solution of minus 30 degrees, which is outside the range.
  4. Add 180 degrees to bring it into range: x = 150 degrees, and adding 180 again gives x = 330 degrees.
  5. The full solution set in the range is 30, 150, 210 and 330 degrees.
How the marks are awarded. 1 mark for tan x = plus or minus 1 over root 3. 1 mark for including both signs. 1 mark for 30 and 210 degrees. 1 mark for 150 degrees. 1 mark for 330 degrees.
Where students lose the mark. Taking only the positive square root, which loses half the solutions. A squared trigonometric function always yields both signs.
Question 3[3 marks]

Prove the identity (1 minus cos2 theta) divided by sin theta equals sin theta.

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Answer: Proved using the Pythagorean identity.
  1. Start with the left hand side and use sin2 theta + cos2 theta = 1.
  2. Rearranged, 1 minus cos2 theta = sin2 theta.
  3. Substituting gives sin2 theta divided by sin theta.
  4. Cancelling one factor of sin theta leaves sin theta, which equals the right hand side, so the identity is proved.
How the marks are awarded. 1 mark for quoting the Pythagorean identity. 1 mark for replacing the numerator with sin2 theta. 1 mark for cancelling to reach sin theta with a concluding statement.
Where students lose the mark. Working on both sides at once. Start from one side and transform it until it matches the other, then state that the identity is proved.
Question 4[5 marks]

Solve 2 cos2 x + cos x minus 1 = 0 for 0 degrees less than or equal to x, less than or equal to 360 degrees.

Show the worked answer
Answer: x = 60, 180 or 300 degrees
  1. Treat it as a quadratic in cos x. Substituting y = cos x gives 2y2 + y minus 1 = 0.
  2. Factorise: (2y minus 1)(y + 1) = 0, so y = 0.5 or y = minus 1.
  3. For cos x = 0.5: the first solution is 60 degrees, and cosine is also positive in the fourth quadrant, giving 360 minus 60 = 300 degrees.
  4. For cos x = minus 1: the only solution in the range is x = 180 degrees.
  5. The full solution set is 60, 180 and 300 degrees.
How the marks are awarded. 1 mark for recognising the quadratic form. 1 mark for factorising. 1 mark for both values of cos x. 1 mark for 60 and 300 degrees. 1 mark for 180 degrees.
Where students lose the mark. Rejecting cos x = minus 1 as impossible. Cosine can equal minus 1 exactly, at 180 degrees. Only values outside minus 1 to 1 are rejected.

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Trigonometry FAQs

How do I find all solutions of a trigonometric equation?

Use the inverse function to find the first solution, then apply the symmetry of the graph to find the others in the stated range. Sine is positive in the first and second quadrants, cosine in the first and fourth, and tangent repeats every 180 degrees.

How do I solve a quadratic in sine or cosine?

Substitute a single letter for the trigonometric function so the equation becomes a standard quadratic. Solve it, then replace the letter and solve each resulting trigonometric equation separately across the given range.

When should I reject a solution?

Reject any value where sine or cosine would need to exceed 1 or fall below minus 1, since those are impossible. Values of exactly plus or minus 1 are valid. Also reject any solution that falls outside the range stated in the question.

How do I prove a trigonometric identity?

Start from one side, usually the more complicated one, and transform it step by step using known identities until it matches the other side. Never manipulate both sides at once, and finish with a statement that the identity is proved.

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Related IGCSE Additional Mathematics topics

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Written to the published Cambridge IGCSE Additional Mathematics (0606) syllabus. Check your school entry code and syllabus year, because Core and Extended candidates are assessed on different content. Last reviewed 2026-08-12.