设x∈[0,π/2]，求函数y=3sinxcosx-3√3(sinx)^2的最大值、最小值，并写出相应x值

y=3sinxcosx-3√3(sinx)^2
=(3/2)sin2x-3√3·[√(（1-cos2x）/2)]^2
=(3/2)sin2x-3√3·[（1-cos2x）/2]
=(3/2)(sin2x+√3·cos2x-√3)
=(3/2){√(1+(√3)^2)[sin2x/√(1+(√3)^2)+√3·cos2x/√(1+(√3)^2)]-√3}
=3[(1/2)·sin2x/2+(√3/2)·cos2x]-(3√3)/2
=3[sin(π/6)·sin2x/2+cos(π/6)·cos2x]-(3√3)/2
=3·cos(2x-π/6)-(3√3)/2
当2x-π/6=2kπ时,即x=kπ+π/12时,y取得最大值:3-(3√3)/2;
当2x-π/6=2kπ+π时,即x=kπ+7π/12时,y取得最大值:-3-(3√3)/2

化简：
y=3/2*sin2x-3√3[（1-cos2x）/2]=3*(1/2*sin2x+√3/2*cos2x)-3√3/2
=3*sin(2x+π/3)-3√3/2
后略