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JEE Advanced 2020 Paper 1 · Mathematics · Trigonometry and Inverse Trigonometric Functions · medium

Consider the equation sin x+sin 2x+sin 3x=0 for x∈[0,2π].

One or more options correct

Consider the equation sin⁡x+sin⁡2x+sin⁡3x=0\sin x+\sin 2x+\sin 3x=0 for x∈[0,2π]x\in[0,2\pi]. Which of the following is/are correct?
  1. A. The equation has exactly 88 solutions in [0,2π][0,2\pi]
  2. B. The sum of all solutions in [0,2π][0,2\pi] is 7π7\pi
  3. C. The equation has a solution in the open interval (π2,2π3)\left(\dfrac{\pi}{2},\dfrac{2\pi}{3}\right)
  4. D. The equation has exactly 44 solutions in the open interval (0,π)(0,\pi)
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Answer: B

Grouping sin⁡x+sin⁡3x=2sin⁡2xcos⁡x\sin x+\sin3x=2\sin2x\cos x, the equation becomes sin⁡2x (2cos⁡x+1)=0\sin2x\,(2\cos x+1)=0. Then sin⁡2x=0\sin2x=0 gives x=0,π2,π,3π2,2πx=0,\tfrac\pi2,\pi,\tfrac{3\pi}2,2\pi, and cos⁡x=−12\cos x=-\tfrac12 gives x=2π3,4π3x=\tfrac{2\pi}3,\tfrac{4\pi}3. So there are 77 solutions in [0,2π][0,2\pi], and (A) with 88 is false. Their sum is (0+π2+π+3π2+2π)+(2π3+4π3)=5π+2π=7π(0+\tfrac\pi2+\pi+\tfrac{3\pi}2+2\pi)+(\tfrac{2\pi}3+\tfrac{4\pi}3)=5\pi+2\pi=7\pi; (B) is true. (C): the two nearest solutions are π2\tfrac\pi2 and 2π3\tfrac{2\pi}3, which are the endpoints, so none lies strictly inside; false. (D): in (0,π)(0,\pi) only π2\tfrac\pi2 and 2π3\tfrac{2\pi}3 occur, i.e. 22 solutions, not 44; false.
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