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TMUA 2016 · Paper 1 · Question 8 of 20

TMUA 2016 Paper 1 Question 8

Trigonometry — Quadratic in sin · repeated root and a double angle. Try it first; the answer and a full worked solution are below.

TMUA 2016 · Paper 1TrigonometryQuadratic in sin · repeated root and a double angle8 options
Find the maximum angle x in the range 0∘ ≤ x ≤ 360∘ which satisfies the equation cos2(2x) + √3sin(2x) − 74 = 0
  1. A30∘
  2. B60∘
  3. C120∘
  4. D150∘
  5. E210∘
  6. F240∘
  7. G300∘
  8. H330∘
Show the answer and worked solution
answer · F
  1. A30∘
  2. B60∘
  3. C120∘
  4. D150∘
  5. E210∘
  6. F240∘
  7. G300∘
  8. H330∘
Replace cos2(2x) by 1 − sin2(2x) to get everything in terms of s = sin(2x): 1 − s2 + √3 s − 74 = 0, so s2 − √3 s + 34 = 0, which is (s − √32)2 = 0. So sin(2x) = √32 is the only possibility. As x runs over 0∘ to 360∘, 2x runs over 0∘ to 720∘, giving 2x = 60∘,  120∘,  420∘,  480∘ and hence x = 30∘,  60∘,  210∘,  240∘. The largest is 240∘; forgetting to double the range for 2x loses the two answers above 180∘.