题目

已知函数f(x)=根号3倍sinx/2cosx/2+cos^2x/2+1/2求f(x)单调减区间

提问时间：2020-07-21

答案

f(x)=√3sin(x/2)cos(x/2)+cos²(x/2)+1/2
=√3/2sinx+(cosx+1)/2+1/2
=√3/2sinx+1/2cosx+1
=sin(x+π/6)+1
令2kπ+π/2≤x+π/6≤2kπ+3π/2 (k∈N)
则2kπ+π/3≤x≤2kπ+4π/3 (k∈N)
∴f(x)的单调递减区间为[2kπ+π/3,2kπ+4π/3](k∈N)