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NDA 2017 Maths · Mathematics · Trigonometric Equations · hard

The number of values of x in [0,π] satisfying sin x+sin 3x=sin 2x is

Single correct option

The number of values of xx in [0,π][0,\pi] satisfying sin⁡x+sin⁡3x=sin⁡2x\sin x+\sin3x=\sin2x is
  1. A. 4
  2. B. 2
  3. C. 3
  4. D. 5
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Answer: (A) 4

sin⁡x+sin⁡3x=2sin⁡2xcos⁡x\sin x+\sin3x=2\sin2x\cos x, so sin⁡2x(2cos⁡x−1)=0\sin2x(2\cos x-1)=0. sin⁡2x=0\sin2x=0 gives x=0,π2,πx=0,\frac\pi2,\pi; cos⁡x=12\cos x=\frac12 gives x=π3x=\frac\pi3. Total 4. Cancelling sin⁡2x\sin2x would lose three of them.
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Original question written in the NDA 2017 Maths pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake