∵y=(√3/3)x=(tanα)x.∴tanα=√3/3.α=kπ+π/6.k∈Z,sin(kπ+π/6)=sinπ/6=1/2 k=2n,n∈N(0,1,2,3,...) ;=-sinπ/6=-1/2,k=2n+1,n∈N (0,1,2,3,...)tanα=tan(kπ+π/6)=tanπ/6=√3/3 ,k∈Z.