JEE Advanced 2021 Paper 2 · Mathematics · Trigonometry and Inverse Trigonometric Functions · medium

Find the number of values of x∈[0,2π] satisfying sin x+sin 2x+sin 3x=cos x+cos 2x+cos 3x.

Numerical answer

Find the number of values of

x\in[0,2\pi]

satisfying

\sin x+\sin 2x+\sin 3x=\cos x+\cos 2x+\cos 3x

Show answer and solution

Answer: 6

Group the terms:

\sin x+\sin3x=2\sin2x\cos x

and

\cos x+\cos3x=2\cos2x\cos x

, so the equation becomes

\sin2x\,(1+2\cos x)=\cos2x\,(1+2\cos x)

, i.e.

(1+2\cos x)(\sin2x-\cos2x)=0

. Case 1:

\cos x=-\frac12

gives

x=\frac{2\pi}3,\frac{4\pi}3

(two solutions). Case 2:

\tan2x=1

gives

2x=\frac\pi4+k\pi

, so

x=\frac\pi8+\frac{k\pi}{2}

with

k=0,1,2,3

[0,2\pi]

(four solutions); the endpoints do not give extra roots. None of these coincide with Case 1, so the total is

6

. A numerical sign-change count on a fine grid confirms

6

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