fx=sin²wx+根号3倍的sinwxsin(wx+π/2)（w＞0）的最小正周期为π

题目

fx=sin²wx+根号3倍的sinwxsin(wx+π/2)（w＞0）的最小正周期为π
求函数fx在区间【0,三分之二π】上的取值范围

答案

fx=sin²wx+√3sinwxsin(wx+π/2)
=sin²wx+√3sinwx*coswx
=sin²wx+√3/2*sin2wx
=(1-cos2wx)/2+√3/2*sin2wx
=√3/2*sin2wx-1/2*cos2wx+1/2
=sin2wxcosπ/6-cos2wx*sinπ/6+1/2
=sin(2wx-π/6)+1/2
最小正周期=2π/2w=π
w=1
f(x)=sin(2x-π/6)+1/2
x∈[0,2π/3]
2x-π/6∈[-π/6,7π/6]
sin(2x-π/6)∈[-1/2,1]
f(x)∈[0,3/2]
取值范围∈[0,3/2]