设函数f(x)=根号3 cos^2x+sinxcosx - 根号3/2 (1)求函数的最小正周期T,并求出函数的单调递增

f(x)=根号3 cos^2x+sinxcosx - 根号3/2
=(√3/2)(1+cos2x)+(1/2)sin2x-√3/2
=(√3/2)cos2x+(1/2)sin2x
=sin(2x+π/3)
(1) 最小正周期T=2π/2=π
单增区间2x+π/3∈[2kπ-π/2,2kπ+π/2]
即x∈[kπ-5π/12,kπ+π/12]
(2) x∈[0,3π] 2x+π/3∈[π/3,6π+π/3]
满足最大值条件的有2x+π/3=π/2,2π+π/2,4π+π/2,
所以x=π/12,π+π/12,2π+π/12,
所有x的和=π/12+π+π/12+2π+π/12=3π+π/4=13π/4