If `cosx+cosy-cos(x+y)=3/2,` then (a)`x+y=0` (b) `x=2y` (c)`x=y` (d) `2x=y`

Explanation

`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`

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