JEE Advanced 2018 Paper 1 · Mathematics · Trigonometry and Inverse Trigonometric Functions · medium

Let S be the set of all x∈[0,2π] satisfying sin x+sin 2x+sin 3x=cos x+cos 2x+cos 3x. Find the sum of all elements of S divided by π.

Numerical answer

Let

S

be the set of all

x\in[0,2\pi]

satisfying

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

. Find the sum of all elements of

S

divided by

\pi

. 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=-\frac12

gives

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

with sum

2\pi

. Case 2:

\tan2x=1

gives

2x=\frac\pi4+k\pi

, so

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

for

k=0,1,2,3

[0,2\pi]

, with sum

\frac{4\pi}8+\frac\pi2(6)=\frac{7\pi}2

. The two cases give distinct solutions, and a numerical sign-change scan finds exactly six roots. The total is

2\pi+\frac{7\pi}2=\frac{11\pi}2

, so the ratio is

5.50

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