题目

函数f(x)=2cosxsin(x+π/3)-根号3sin^2x+sinxcosx的最小值为多少?此时x等于多少

提问时间：2020-10-09

答案

f(x)=2cosxsin(x+π/3)-√3sin^2x+sinxcosx
=2cosx(√3/2sinx+1/2cosx)-√3sin^2x+1/2sin2x
=√3/2sin2x+cos^2x-√3sin^2x+1/2sin2x
=(√3+1)/2sin2x+cos^2x-√3sin^2x
=(√3+1)/2sin2x+(cos2x+1)/2+√3/2(cos2x-1)
=(√3+1)/2(sin2x+cos2x)+(√3-1)/2
=(√6+√2)/2sin(2x+π/4)+(√3-1)/2
2x+π/4=2kπ-π/2 , x=kπ-3π/8 , k∈Z
最小值f(kπ-3π/8)=-(√6+√2)/2+(√3-1)/2