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
則f(x)=sin(2x+π/6)
單調(diào)增區(qū)域?yàn)椋?π/2+2kπ≤2x+π/6≤π/2+2kπ,k∈Z
即-π/3+kπ≤x≤π/6+kπ,k∈Z
∴f(x)的單調(diào)遞增區(qū)間為：[-π/3+kπ,π/6+kπ],k∈Z
2、當(dāng)x∈[0,π/2]時(shí),
2x+π/6∈[π/6,7π/6]
畫出草圖圖像可知：
函數(shù)f(x)的值域?yàn)椋篬-1/2,1]