已知函数f(x)=(√3/2 ) sinwx- sin²(wx/2)+1/2的最小正周期为派.急

1、f(x)=(√3/2 ) sinwx- sin²(wx/2)+1/2
=(√3/2 ) sinwx+1/2[-2sin²(wx/2)+1]
=(√3/2 ) sinwx+1/2coswx
=sin(wx+π/6)
∵最小正周期：T=2π/w=π
∴x=2
则f(x)=sin(2x+π/6)
单调增区域为：-π/2+2kπ≤2x+π/6≤π/2+2kπ,k∈Z
即-π/3+kπ≤x≤π/6+kπ,k∈Z
∴f(x)的单调递增区间为：[-π/3+kπ,π/6+kπ],k∈Z
2、当x∈[0,π/2]时,
2x+π/6∈[π/6,7π/6]
画出草图图像可知：
函数f(x)的值域为：[-1/2,1]