Question

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区间

（2）求在【0,3派）内使函数取得最大值的所有X的和
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Answer 1

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
希望能帮到你，祝学习进步O(∩_∩)O

Answer 2

f(x)=根号3 cos^2x+sinxcosx - 根号3/2
=根号3cos2x/2+sin2x/2
=sin(2x+π/3)
所以最小正周期为T=2π/2=π
单调递增区间为[nπ-5π/12,nπ+π/12]
单调递减区间为[nπ+π/12,nπ+7π/12]

在[0,3π)之间，取最大值的时候，x的解为x=π/12，x=13π/12，x=25π/12
求和得：13π/4

Answer 3

f(x)=√3/2*[2(cosx)^2-1+1/2(2sinxcosx)
=1/2*sin2x+√3/2*cos2x
=sin2x*cosπ/3+cos2x*sinπ/3
=sin(2x+π/3)
所以最小正周期为
2π/2=π
单调增区间为-π/2+2kπ<=2x+π/3<=π/2+2kπ
得到-5/12π+kπ<=x<=π/12+kπ
【0,3派）内x值有
π/12,13/12π,25/12π
和为39/12π