For the smallest positive values of `xa n dy ,` the equation `2(s inx+s in y)-2cos(x-y)=3` has a solution, then which of the following is/are true? `sin(x+y)/2=1` (b) `cos((x-y)/2)=1/2` number of ordered pairs `(x , y)` is 2 number of ordered pairs `(x , y)i s3`

Options

A. `"sin" (x+y)/2=1`
B. `cos((x-y)/2)=1/2`
C. number of ordered pairs (x, y) is 2
D. number of ordered pairs (x, y) is 3

Answer

A, B, C

Explanation

The given equation is
`2(sin x+ sin y)-2 cos(x-y)=3`
`rArr 2xx2 "sin" (x+y)/2"cos" (x-y)/2-2[2 "cos"^(2) (x-y)/2-1]=3`
or `4 cos^(2) ((x-y)/2)-4 sin ((x+y)/2)cos((x-y)/2)+1=0`
or `cos((x-y)/2)=(4 sin((x+y)/2) pm sqrt(16 sin^(2) ((x+y)/2)-16))/8`
`:. sin^(2) ((x+y)/2) ge 1`
`rArr "sin" (x+y)/2= pm 1`
Since x and y are smallest and positive, we have
`"sin" (x+y)/2=1` and `(x+y)/2=pi/2`
i.e., `x+y=pi` ...(i)
Also, `cos ((x-y)/2)=1/2`
`rArr x-y=(2pi)/3 or - (2pi)/3` ...(ii)
From Eqs. (i) and (ii), we get
`(x=5pi//6, y=pi//6) or (x=pi//6, y=5pi//6)`