The Chief Examiner reported that this question was attempted poorly by majority of the candidates. According to the report, they showed poor understanding of trigonometric ratios. Candidates were expected to recall that cos 2θ = cos2θ – sin2θ = (1 – sin2θ) – sin2θ = 1 - 2 sin2θ. Hence, sinθ + cos 2θ = 0 becomes sinθ + 1 - 2 sin2θ = 0 which is equivalent to

2 sin2θ – sinθ - 1 = 0. By substituting y for sinθ, the equation becomes 2y2 – y – 1 = 0. Solving this quadratic equation correctly gave y = sinθ = 1 or . This implied that θ = 90o, 210o or 330o. Hence, the truth set was {90o, 210o, 330o}.