已知2cosxcos（x-π/6）-根号三sin^2x+sinxcosx求f（x）的最小正周期 当x属于【0,π】时,若f（x）=1

f(x)=2cosxcos(x-π/6)-√3sin^2x+sinxcosx =2cosxcos(x-π/6)-√3sin^2x+sinxcosx=2cosx(√3/2cosx+1/2sinx)-√3in^2x+sinxcosx=√3cos^2x+sinxcosx-√3sin^2x+sinxcosx=√3(cos^2x-sin^2x)+2sinxcosx=√3cos2x+sin...