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JEE Main 2017 · Mathematics · Trigonometry and Inverse Trigonometric Functions · medium

The number of solutions of the equation sin x+sin 2x+sin 3x=0 in the interval [0,2π] is

Single correct option

The number of solutions of the equation sin⁡x+sin⁡2x+sin⁡3x=0\sin x+\sin2x+\sin3x=0 in the interval [0,2π][0,2\pi] is
  1. A. 55
  2. B. 66
  3. C. 77
  4. D. 88
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Answer: (C) 77

Group sin⁡x+sin⁡3x=2sin⁡2xcos⁡x\sin x+\sin3x=2\sin2x\cos x, so the equation becomes sin⁡2x (2cos⁡x+1)=0\sin2x\,(2\cos x+1)=0. Case 1: sin⁡2x=0\sin2x=0 gives x=0,π2,π,3π2,2πx=0,\tfrac\pi2,\pi,\tfrac{3\pi}2,2\pi, i.e. 55 solutions. Case 2: cos⁡x=−12\cos x=-\tfrac12 gives x=2π3,4π3x=\tfrac{2\pi}3,\tfrac{4\pi}3, i.e. 22 solutions. The two cases share no common roots, so the total is 5+2=75+2=7. Dropping the endpoint 2π2\pi or one case is the common mistake.
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