The number of ordered pairs which satisfy the equation `x^2+2xsin(x y)+1=0` are (where `y in [0,2pi]` ) (a) 1 (b) 2 (c) 3 (d) 0

Explanation

`x^2+2xsin(xy) +1 = 0`

`=>(x+sin(xy))^2 - sin^2(xy)+1 = 0`

`=>(x+sin(xy))^2 + cos^2(xy) = 0`

`=>x+sin(xy) = 0 and cos(xy) = 0`

When, `cos(xy) = 0, sin(xy) = sqrt(1-cos^2(xy)) = +-1`

When `sin(xy) = 1`

`x+sin(xy) = 0=>x+1 = 0 => x = -1`

`=>sin(-y) = 1 =>siny = -1 => y = (3pi)/2`

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