高一数学题(急求详细答案,带解析的）

f(x)=5sinxcosx-5根号3cos平方x+5/2根号3
=5(1/2 *sin2x-√3cos²x+√3/2)
=5[1/2 *sin2x-√3/2*(2cos²x -1)]
=5(1/2 sin2x-√3/2*cos2x)
=5sin(2x- π/3)
(1)周期T=2π/2 =π
(2)由-π/2 +2kπ《2x- π/3《π/2 +2kπ
得-π/12 +kπ《x《5π/12 +kπ
∴函数的单调增区间是[-π/12 +kπ,5π/12 +kπ] k∈Z
由π/2 +2kπ《2x- π/3《3π/2 +2kπ
得由5π/12+ kπ《x《11π/12 +kπ
∴函数的单调减区间是[5π/12 +kπ,11π/12 +kπ] k∈Z
(3)由2x- π/3=π/2 +kπ 得函数的对称轴方程 x=5π/12 +kπ/2
由2x- π/3=kπ 得x=π/6+kπ/2 ∴对称中心坐标是(π/6+kπ/2 ,0) k∈Z