f(x)=sinx-cos(x+π/6)=sinx-(cosxcosπ/6-sinxsinπ/6)=sinx-[(√3/2)cosx-(1/2)sinx]=(3/2)sinx-(√3/2)cosx=√3[(√3/2)sinx-(1/2)cosx]=√3[cosπ/6sinx-sinπ/6cosx]=√3sin(x-π/6)值域为 [-√3,√3]不懂继续...