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If 0 le x le 2pi, then the number of rea...

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

A

5

B

7

C

9

D

3

Text Solution

Verified by Experts

The correct Answer is:
B
`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`
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Knowledge Check

  • If 0 le x lt 2pi , then the number of real values of x which satisfy the equation cosx+cos2x+cos3x+cos4x=0 , is :

    A
    3
    B
    5
    C
    7
    D
    9

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    A
    1
    B
    2
    C
    3
    D
    4

  • Number of values of x in [0, 2pi] and satisfying the equation cos x cos2x cos3x=1//4 is :

    A
    6
    B
    8
    C
    20
    D
    24

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