If `0 le x le 2pi`, then the number of real values of x, which satisfy the equation `cos x + cos 2x + cos 3x + cos 4x=0`, is

Options

A. 5
B. 7
C. 9
D. 3

Answer

B

Explanation

`cos x + cos 2x + cos 3x + cos 4x=0, 0 le x lt 2pi`
`rArr (cos x + cos 4x) + (cos 2x + cos 3x)=0`
`rArr 2 "cos"(5x)/2 "cos" (3x)/2+2 cos (5 x)/2 "cos" x/2 =0`
`rArr 2 "cos" (5x)/2 [2 cos x cos x/2]=0`
`rArr "cos" (5x)/2 =0`
or `cos x=0` or `"cos" x/2 =0`
`rArr x=((2n+1)pi)/5` or `x=(2n+1) pi/2`
or `x=(2n+1) pi, n in Z`
`rArr x={pi/5, (3pi)/5, pi, (7pi)/5, (9pi)/5, pi/2, (3pi)/2}`
`:.` Number of solution is `7`