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JEE Advanced 2022 Paper 1 · Mathematics · Trigonometry and Inverse Trigonometric Functions · medium

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

One or more options correct

Consider the equation sin⁡x+sin⁡2x+sin⁡3x=cos⁡x+cos⁡2x+cos⁡3x\sin x+\sin2x+\sin3x=\cos x+\cos2x+\cos3x for x∈[0,2π]x\in[0,2\pi]. Which of the following statements is/are correct?
  1. A. The equation has exactly six solutions in [0,2π][0,2\pi].
  2. B. The sum of all solutions in [0,2π][0,2\pi] is 6π6\pi.
  3. C. Exactly three solutions lie in the open interval (0,π)(0,\pi).
  4. D. x=π2x=\dfrac{\pi}{2} is a solution.
Show answer and solution

Answer: A, C

Grouping, sin⁡2x(1+2cos⁡x)=cos⁡2x(1+2cos⁡x)\sin2x(1+2\cos x)=\cos2x(1+2\cos x), i.e. (1+2cos⁡x)(sin⁡2x−cos⁡2x)=0(1+2\cos x)(\sin2x-\cos2x)=0. Then cos⁡x=−12\cos x=-\frac12 gives x=2π3,4π3x=\frac{2\pi}3,\frac{4\pi}3, and tan⁡2x=1\tan2x=1 gives x=π8,5π8,9π8,13π8x=\frac\pi8,\frac{5\pi}8,\frac{9\pi}8,\frac{13\pi}8. (A) true: six solutions. (B) false: the sum is 2π+28π8=11π22\pi+\frac{28\pi}{8}=\frac{11\pi}2. (C) true: π8,5π8,2π3\frac\pi8,\frac{5\pi}8,\frac{2\pi}3 lie in (0,π)(0,\pi). (D) false: at x=π2x=\frac\pi2 the left side is 1+0−1=01+0-1=0 while the right side is 0−1+0=−10-1+0=-1.
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