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A-LevelMathematicsTrigonometry9709/12 · Feb/Mar 2020 · Q55 Marks

Solve the equation tan θ + 3 sin θ + 2 = 2 tan θ - 3 sin θ + 1 for 0° ≤ θ ≤ 90°.

  1. Solve the equation tan θ + 3 sin θ + 2 = 2 tan θ - 3 sin θ + 1 for 0° ≤ θ ≤ 90°.[5]

Mark scheme

Marking points from the official mark scheme (Feb/Mar 2020). The mark or mark type is shown in brackets.

  1. [M1] tan θ + 3 sin θ + 2 = 2(tan θ - 3 sin θ + 1) => 2 tan θ – 6 sin θ + 2 = tan θ + 3 sin θ + 2 → tan θ – 9 sin θ (= 0) Multiply by denominator and simplify
  2. [M1] sin θ/cos θ – 9 sin θ (= 0) Multiply by cos θ
  3. [M1] sin θ (1/cos θ – 9) (= 0) → sin θ = 0, cos θ = 1/9 Factorise and attempt to solve at least one of the factors = 0
  4. [A1] θ = 0
  5. [A1] 83.6° (only answers in the given range)

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About this question

Topic
Trigonometry, A-Level Mathematics (syllabus 9709). Structured question, 5 marks.
Source
A-Level Mathematics, Feb/Mar 2020 Paper 1 Variant 2, question 5.

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