已知函数sinx^2+2√3sinxcosx+sin(x+π⼀4)sin(x-π⼀4) x∈R

f(x)=(sinx)^2+2√3sinxcosx+sin(x+π/4)sin(x-π/4)
=1/2(1-cos2x)+√3sin2x+sin(x+π/4)sin[-π/2+(x+π/4)]
=1/2-1/2cos2x+√3sin2x-sin(x+π/4)cos(x+π/4)
=1/2-1/2cos2x+√3sin2x-1/2sin(2x+π/2)
=1/2-1/2cos2x+√3sin2x-1/2cos2x
=√3sin2x-cos2x+1/2
=2(√3/2sin2x-1/2cos2x)+1/2
=2sin(2x-π/6)+1/2
f(x)最小正周期T=2π/2=π
f(x)max=2+1/2=5/2
f(x)min=-2+1/2=-3/2
f(x)值域为[-3/2,5/2]
(2)
x=x0(0≤x0≤π/2)为f(x)的一个零点,

则f(x0)=2sin(2x0-π/6)+1/2=0
∴sin(2x0-π/6)=-1/4<0
∵0≤x0≤π/2
∴-π/6≤2x0-π/6≤5π/6
∴-π/6≤2x0-π/6<0
∴cos(2x0-π/6)=√[1-sin²(2x0-π/6)]=√15/4
∴sin2x0
=sin[(2x0-π/6)+π/6]
=sin(2x0-π/6)cosπ/6+cos(2x0-π/6)sinπ/6
=-1/4*√3/2+√15/4*1/2
=(√15-√3)/8