JEE Advanced 2025 Paper 2 · Mathematics · Trigonometry and Inverse Trigonometric Functions · medium

Let S be the sum of all solutions of sin x+sin 2x+sin 3x=cos x+cos 2x+cos 3x in the interval [0,2π]. If S=kπ, find k.

Numerical answer

Let

S

be the sum of all solutions of

\sin x+\sin2x+\sin3x=\cos x+\cos2x+\cos3x

in the interval

[0,2\pi]

. If

S=k\pi

, find

k

. Give the answer correct to two decimal places.

Show answer and solution

Answer: 5.50

Group the terms:

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

and

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

. The equation becomes

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

, i.e.

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

. Case 1:

\cos x=-\tfrac12

gives

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

(sum

2\pi

). Case 2:

\tan2x=1

gives

2x=\frac\pi4+m\pi

, so

x=\frac\pi8,\frac{5\pi}8,\frac{9\pi}8,\frac{13\pi}8

(sum

\frac{28\pi}{8}=3.5\pi

). No overlap, so six solutions, confirmed numerically, and

S=5.5\pi

; hence

k=5.50

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