JEE Advanced 2024 Paper 1 · Mathematics · Trigonometry and Inverse Trigonometric Functions · medium

The number of solutions of the equation sin x+sin 3x+sin 5x=cos x+cos 3x+cos 5x in the interval [0,2π] is ______.

Numerical answer

The number of solutions of the equation

\sin x+\sin 3x+\sin 5x=\cos x+\cos 3x+\cos 5x

in the interval

[0,2\pi]

is ______.

Show answer and solution

Answer: 10

Group the terms:

\sin x+\sin5x=2\sin3x\cos2x

and

\cos x+\cos5x=2\cos3x\cos2x

. The equation becomes

\sin3x(1+2\cos2x)=\cos3x(1+2\cos2x)

, that is

(\sin3x-\cos3x)(1+2\cos2x)=0

. Case 1:

\tan3x=1

(with

\cos 3x\neq0

), so

3x=\tfrac\pi4+k\pi

; as

3x\in[0,6\pi]

k=0,1,\dots,5

, giving

6

values of

x

. Case 2:

\cos2x=-\tfrac12

, so

2x\in\{\tfrac{2\pi}{3},\tfrac{4\pi}{3},\tfrac{8\pi}{3},\tfrac{10\pi}{3}\}

[0,4\pi]

, giving

4

values of

x

(

\tfrac\pi3,\tfrac{2\pi}3,\tfrac{4\pi}3,\tfrac{5\pi}3

). No value belongs to both families: for

\cos2x=-\tfrac12

one has

3x\equiv\pm\pi,\ 0 \pmod{2\pi}

-type angles, where

\tan3x

0

rather than

1

, so the two lists are disjoint and the

10

roots are distinct. The total is

6+4=10

, confirmed by a numerical sign-change scan.

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