函数f(x)=√3sin^(wx/2)+sin(wx/2)cos(wx/2) (w>0)的周期为π,求w的值和函数f(x

f(x)=√3sin²(wx/2)+sin(wx/2)cos(wx/2)
=-(√3/2)*[1- 2sin²(wx/2) -1] +(1/2)*2sin(wx/2)cos(wx/2)
=-(√3/2)*[cos(wx) -1]+(1/2)*sin(wx)
=sin(wx)*cos(π/3) -cos(wx)*sin(π/3)+ (√3/2)
=sin(wx - π/3)+ (√3/2)
由已知得：函数周期T=2π/w=π,解得：w=2
那么函数解析式可写为：
f(x)=sin(2x - π/3)+ (√3/2)
当2x- π/3 ∈[-π/2 +2kπ,π/2 +2kπ],即x∈[-π/12 +kπ,5π/6 +kπ],k∈Z时,函数f（x）是增函数,所以函数的单调递增区间为[-π/12 +kπ,5π/6 +kπ],k∈Z