If `cos(x+pi/3)+cos x=a`has real solutions, thennumber of integral values of `a`are 3sum of number of integral values of `ai s0`when `a=1`, number of solutions for `x in [0,2pi]`are 3when `a=1,`number of solutions for `x in [0,2pi]`are 2A. number of integral values of a are 3B. sum of number of integral values of a is 0C. when `a=1`, number of solution for `x in [0, 2pi]` are 3D. when `a=1`, number of solutions for `x in [0, 2pi]` are 2

If `cos(x+pi/3)+cos x=a`has real solutions, thennumber of integral val

1.	If `cos(x+pi/3)+cos x=a`has real solutions, thennumber of integral values of `a`are 3sum of number of integral values of `ai s0`when `a=1`, number of solutions for `x in [0,2pi]`are 3when `a=1,`number of solutions for `x in [0,2pi]`are 2A. number of integral values of a are 3B. sum of number of integral values of a is 0C. when `a=1`, number of solution for `x in [0, 2pi]` are 3D. when `a=1`, number of solutions for `x in [0, 2pi]` are 2
Answer» Correct Answer - A::B::D `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))`.