The number of solution of `sin x+sin 2x+sin 3x`
`=cos x +cos 2x+cos 3x, 0 le x le 2pi`, is

Options

A. 7
B. 5
C. 4
D. 6

Answer

D

Explanation

We have
`(sin x+ sin 3x)+sin 2x=(cos x+cos 3x)+cos 2x`
or `2 sin 2x cos x+ sin 2x=2 cos 2x cos x + cos 2x`
or `sin 2x(2 cos x +1)=cos2x(2 cso x +1)`
or `(2 cos x+1) (sin 2x- cos 2x)=0`
or `cos x=-1//2 or sin 2x-cos 2x=0`
`rArr x=2 n pi pm (2 pi//3) or tan 2x=1=tan (pi//4)`
`rArr x=2n pi pm (2pi//3) or x=(4n+1) pi//8, n in Z`
But here `0 le x le 2pi`. Hence, `x=pi//8, 5pi//8, 2pi//3, 9 pi//8, 4pi//3, 13 pi//8`.