已知f(x)=sin(2x+π/3)-根号3sin^2x+sinxcosx+根号3/2

题目

已知f(x)=sin(2x+π/3)-根号3sin^2x+sinxcosx+根号3/2
①求函数的最小正周期②求函数的最小值及此时x的值③求函数的单调增区间

答案

由题意可得:
f(x)=sin(2x+π/3)-√3sin^2x+sinxcosx+√3/2
=sin(2x+π/3)-√3(1/2-1/2cos2x)+1/2sin2x+√3/2
=2sin(2x+π/3)
(1)最小正周期2π/2=π
(2)函数的最小值为-2,此时x=kπ-5π/12
(3)函数的单调增区间为[kπ-5π/12,kπ+π/12]