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JEE Advanced 2022 Paper 2 · Mathematics · Trigonometry and Inverse Trigonometric Functions · medium

The number of values of x∈[0,2π] for which sin x+sin 2x+sin 3x=0 but cos x+cos 2x+cos 3x≠0 is

Numerical answer

The number of values of x∈[0,2π]x\in[0,2\pi] for which sin⁡x+sin⁡2x+sin⁡3x=0\sin x+\sin 2x+\sin 3x=0 but cos⁡x+cos⁡2x+cos⁡3x≠0\cos x+\cos 2x+\cos 3x\neq0 is
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Answer: 5

Group the sines: sin⁡x+sin⁡3x=2sin⁡2xcos⁡x\sin x+\sin3x=2\sin2x\cos x, so the sum is sin⁡2x (2cos⁡x+1)\sin2x\,(2\cos x+1). Similarly cos⁡x+cos⁡3x=2cos⁡2xcos⁡x\cos x+\cos3x=2\cos2x\cos x, so the cosine sum is cos⁡2x (2cos⁡x+1)\cos2x\,(2\cos x+1). The sine sum vanishes when sin⁡2x=0\sin2x=0, giving x=0,π2,π,3π2,2πx=0,\frac\pi2,\pi,\frac{3\pi}2,2\pi, or when cos⁡x=−12\cos x=-\frac12, giving x=2π3,4π3x=\frac{2\pi}3,\frac{4\pi}3: seven roots in all. At cos⁡x=−12\cos x=-\frac12 both factors 2cos⁡x+12\cos x+1 vanish, so the cosine sum is also zero and these two are excluded. At the five roots with sin⁡2x=0\sin2x=0 we have cos⁡2x=±1\cos2x=\pm1 and 2cos⁡x+1∈{3,1,−1}2\cos x+1\in\{3,1,-1\}, so the cosine sum is nonzero. The count is 7−2=57-2=5.
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