x+siny=2020 ⋯(1)x+2020cosy=2019 ⋯(2)
From eqns. (1) and (2),
siny−2020cosy=1
⇒siny=1+2020cosy
As 0≤y≤π2 and we know that
siny≤1 and 1+2020cosy≥1
So, they will be equal only when,
siny=1 and cosy=0
⇒y=π2
Using equation (1),
x=2020−siny=2019
[x+y]=[2019+π2] =2019+[π2] =2019+1 =2020