The values of `x_1` between 0 and `2pi` , satisfying the equation `cos3x+cos2x=sin(3x)/2+sinx/2` are `pi/7` (b) `(5pi)/7` (c) `(9pi)/7` (d) `(13pi)/7`

Options

A. `pi/7`
B. `(5 pi)/7`
C. `(9 pi)/7`
D. `(13 pi)/7`

Answer

A, B, C, D

Explanation

`cos 3x+cos 2x = "sin" (3x)/2 + "sin"x/2, 0 lt x lt 2pi`
or `2 "cos" (5pi)/2"cos" x/2=2 sin x " cos" x/2`
or `"cos" x/2["cos" (5pi)/2- sin x]=0`
`rArr "cos" x/2=0 or "cos" (5 x)/2= sin x`
`rArr x/2= 2n pi pm pi/2 or (5pi)/2=2 n pi pm (pi/2-x), n in Z`
`rArr x/2=(4n pm 1) pi/2 or (5x)/2=2 n pi + pi/2-x, (5 x)/2=2n pi -pi/2+x`
`rArr x=(4n pm 1) pi or x= (4 n pi)/7+pi/7, x=4n pi/3-pi/3`
Since, `0 lt x lt 2pi`,
`x=pi/7, (5pi)/7, pi, (9pi)/7, (13 pi)/7`