If `x , y in [0,2pi]` , then find the total number of ordered pairs `(x , y)` satisfying the equation `sinxcosy=1`

Explanation

`sin x cos y=1`
`rArr sin x=1, cos y=1 or sin x=-1, cos y =-1`
If `sin x=1. cos y=1`, hence, `x=pi//2, y=0, 2pi`
If `sin x=-1, cos y =-1`, hence, `x=3pi//2, y=pi`
Thus, the possible ordered ordered pairs are `(pi/2, 0), (pi/2, 2pi)` and `((3pi)/2, pi)`.