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

Options

A. `x+y=0`
B. x=2y
C. x=y
D. 2x=y

Answer

B

Explanation

`cos x+cosy-cos(x+y)=(3)/(2)`
or `2cos((x+y)/(2))cos((x-y)/(2))-2cos^(2)((x+y)/(2))+1=(3)/(2)`
or `2cos^(2)((x+y)/(2))-2cos((x+y)/(2))cos((x-y)/(2))+(1)/(2)=0`.
Now `cos((x+y)/(2))` is always real, then discriminant `ge0`. Thus,
`4cos^(2)((x-y)/(2))-4ge0`
or `cos^(2)((x-y)/(2))ge1`
or `cos^(2)((x-y)/(2))=1`
or `(x-y)/(2))=0` or x=y.