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NDA 2016 Maths · Mathematics · Trigonometric Equations · hard

The number of solutions of the equation sin 2x=sin x in the interval [0,2π) is

Single correct option

The number of solutions of the equation sin⁡2x=sin⁡x\sin2x=\sin x in the interval [0,2π)[0,2\pi) is
  1. A. 44
  2. B. 22
  3. C. 33
  4. D. 66
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Answer: (A) 44

sin⁡2x=sin⁡x\sin2x=\sin x gives sin⁡x(2cos⁡x−1)=0\sin x(2\cos x-1)=0. Then sin⁡x=0\sin x=0 gives x=0,πx=0,\pi, and cos⁡x=12\cos x=\dfrac12 gives x=π3,5π3x=\dfrac\pi3,\dfrac{5\pi}3. That is 4 solutions. Cancelling sin⁡x\sin x from both sides loses the solutions 00 and π\pi and leads to the wrong count 2.
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Original question written in the NDA 2016 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