函数f(x)=[2sin(x+π/3)+sinx]cosx-根3sin^2x,(x∈R).1)求函数f(x)的最小正周期

f(x)=[2sin(x+π/3)+sinx]cosx-√3sin^2x
=[sinx+√3cosx+sinx]cosx-√3sin^2x
=2sinxcosx+√3cos^2x-√3sin^2x
=sin2x+√3cos2x
=2sin(2x+π/3)
f(x)的最小正周期为π
根据题意,m只需大于f(x)在该区间的最小值就行
当x=5π/12时,f(x)取最小值1
所以m>1