If x+y=2π3 and sinxsiny=2 then

x+y=2π3⇒y=2π3−x∴sinx=2sin2π3−x                = 232cosx+12sinx               = 3cosx+sinx⇒cosx=0⇒x=nπ+π2⇒y=2π3−nπ−π2=π6−nπHence,x∈0,4π,x=π2,3π2,5π2,7π2and for  x∈0,4π,y=π6,7π6,13π6,19π6