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NDA 2020 Maths · Mathematics · Trigonometric Equations · hard

The sum of all solutions of sin 3x+sin x=sin 2x lying in [0,π] is

Single correct option

The sum of all solutions of sin⁡3x+sin⁡x=sin⁡2x\sin3x+\sin x=\sin2x lying in [0,π][0,\pi] is
  1. A. 3π2\dfrac{3\pi}{2}
  2. B. 5π3\dfrac{5\pi}{3}
  3. C. 11π6\dfrac{11\pi}{6}
  4. D. 2π2\pi
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Answer: (C) 11π6\dfrac{11\pi}{6}

sin⁡3x+sin⁡x=2sin⁡2xcos⁡x\sin3x+\sin x=2\sin2x\cos x, so sin⁡2x(2cos⁡x−1)=0\sin2x(2\cos x-1)=0. Then sin⁡2x=0\sin2x=0 gives x=0,π/2,πx=0,\pi/2,\pi and cos⁡x=12\cos x=\tfrac12 gives x=π/3x=\pi/3. Sum =0+π2+π+π3=11π6=0+\dfrac\pi2+\pi+\dfrac\pi3=\dfrac{11\pi}{6}.
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Original question written in the NDA 2020 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