`cosx+cosy - cos(x+y) = 3/2`
`=>2cos((x+y)/2)cos((x-y)/2) - (2cos^2(x+y)-1) = 3/2`
`=>2cos^2(x+y) - 2cos((x+y)/2)cos((x-y)/2) +3/2-1 = 0`
`=>2cos^2(x+y) - 2cos((x+y)/2)cos((x-y)/2) +1/2 = 0`
Now, this ia quadratic equation in `cos(x+y)/2`.
So, to have real roots, its discriminant should be greater than or equal to `0`.
`:. (-2cos((x-y)/2))^2 -4(2)(1/2) ge 0`
`=>4cos^2((x-y)/2) - 4 ge 0`
`=>cos^2((x-y)/2) ge 1`
`=>cos((x-y)/2) ge 1`
As `cos theta` can not be greater than `1`.
So, only possible value for the above inequality is `1`.
`:. cos((x-y)/2) = 1`
`=>cos((x-y)/2) = cos 0 `
`=>(x-y)/2 = 0 `
`=> x = y`
So, option `(c)` is the correct option.
