If `cos(x+pi/3)+cos x=a` has real solutions, then number of integral values of `a` are 3 sum of number of integral values of `ai s0` when `a=1` , number of solutions for `x in [0,2pi]` are 3 when `a=1,` number of solutions for `x in [0,2pi]` are 2

Options

A. number of integral values of a are 3
B. sum of number of integral values of a is 0
C. when `a=1`, number of solution for `x in [0, 2pi]` are 3
D. when `a=1`, number of solutions for `x in [0, 2pi]` are 2

Answer

A, B, D

Explanation

`cos (x+pi//3)+cos x=a`
or `1/2 cos x-(sqrt(3)/2) sin x + cos x =a`
or `(3/2) cos x- (sqrt(3)/2) sin x=a`
`rArr -sqrt((9/4+3/4)) le a le sqrt((9/4+3/4))`
`rArr -sqrt(3) le a le sqrt(3)` ...(i)
Hence, there are three integral values of `a=-1, 0, 1` whose sum is 0.
For `a=1`, the given equation is `(sqrt(3)//2) cos x - (1//2) sin x =1//sqrt(3)`.
Thus,
`cos (x+pi/6)=1/sqrt(3)`
`rArr x+pi/6=2n pi pm alpha`, where `alpha=cos^(-1) (1/sqrt(3))`
`rArr x=2npi - pi/6 pm alpha`
Hence, the solution for `a=1` in `[0, 2pi]` are
`cos^(-1) (1//sqrt(3))-11pi//6, 11pi//6-cos^(-1) (1//sqrt(3))`.