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. 3
B. 5
C. 7
D. 9

Answer

C

Explanation

We have,
`"cos" x+ "cos" 2x + "cos" 3x + "cos" 4x = 0`
`rArr ("cos" x + "cos" 3x) +("cos" 2x +"cos" 4x) = 0`
`rArr 2"cos" 2x"cos" x + 2 "cos" 3x "cos" x = 0`
`rArr 2 "cos" x ("cos" 2x + "cos" 3x) = 0`
`rArr 4 "cos" x "cos" (5x)/(2) "cos" (x)/(2) = 0`
`rArr "cos" x = 0" or, cos" (5x)/(2) "or, cos" (x)/(2) = 0`
`rArr x = (pi)/(2), (3pi)/(2), (5pi)/(2) = (pi)/(2), (3pi)/(2), (5pi)/(2), (7pi)/(2), (9pi)/(2),(x)/(2) = (pi)/(2),`
`rArr x = (pi)/(5), (3pi)/(5), (pi)/(2), pi, (7pi)/(5), (3pi)/(2), (7pi)/(5)`