题目

已知函数f（x）=2cosx*sin(x+π/3)-√3sin^2x+sinx*cosx
(1)求函数f(x)的单调递减区间
（2）将函数f(x)的图像按向量a=（m,0)平移后得到g(x)的图像,求使函数g(x)为偶函数的m的最小正值

提问时间：2020-07-24

答案

1.
f（x）=2cosx*sin(x+π/3)-√3sin^2x+sinx*cosx
= 2cosx*sin(x+π/3)- 2sinx*[(√3/2)sinx-(1/2)cosx]
= 2cosx*sin(x+π/3)- 2sinx*[sin(π/3)sinx-cos(π/3)cosx)]
= 2cosx*sin(x+π/3)+ 2sinxcos(x+π/3)
= 2sin(2x+π/3)
f(x)的单调区间
2kπ - π/2
2sin[2(x-m)+π/3] = 2sin{[2(x+m)-π/3]}
=>
2(x-m) + π/3 + 2kπ = 2(x+m)-π/3
=>
4m= 2π/3 + 2kπ
m=π/6 取最小正值(k=0)