f(x)=(sinwx)^2 +√3 sinwxsin(wx+ π/2)
=(sinwx)^2 +√3 sinwxcoswx
=[(1-cos2wx)/2]+(√3 /2)sin2wx
=(√3 /2)sin2wx-(1/2)cos2wx+(1/2)
=sin(2wx- π/6)+(1/2)
則 2π/(2w)=π
解得 w=1
所以 f(x)=sin(2x- π/6)+(1/2)
0≤x≤2π/3時(shí)
0≤2x≤4π/3
-π/6≤2x- π/6≤7π/6
則 -1/2≤sin(2x- π/6)≤1
所以 f(x)的取值范圍是[0,3/2]