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A-LevelMathematicsTrigonometry9709/32 · May/June 2022 · Q25 Marks

Solve the equation 3 cos 2θ = 3 cos θ + 2, for 0° ≤ θ ≤ 360°.

  1. (a)Solve the equation 3 cos 2θ = 3 cos θ + 2, for 0° ≤ θ ≤ 360°.[5]

Mark scheme

Marking points from the official mark scheme (May/June 2022). The mark or mark type is shown in brackets.

  1. M1 Use correct double-angle formula to obtain an equation in cosθ e.g. 3(2cos²θ-1) = 3cosθ + 2
  2. A1 Obtain 6cos²θ – 3cosθ – 5 = 0, or 3-term equivalent M1 A0 is scored if they use any correct formula for cos 2θ and make a subsequent error.
  3. M1 Solve a 3-term quadratic in cosθ for θ As far as θ = cos⁻¹ ((3-√129)/12) if quadratic correct.
  4. A1 Obtain a correct answer, e.g. 134.1° Accept greater accuracy e.g. 134.1456, 225.8544.
  5. A1 FT Obtain a second answer, e.g. 225.9° and no other in [0°, 360°] Treat answers in radians (2.34 and 3.94) as a misread. Ignore answers outside [0°, 360°]. The FT is for 360° minus the first answer.

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About this question

Topic
Trigonometry, A-Level Mathematics (syllabus 9709). Structured question, 5 marks.
Source
A-Level Mathematics, May/June 2022 Paper 3 Variant 2, question 2.

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