Question

已知函数f(x)=sin^2x+根号3sinxcosx,x属于R

求f（x）的最小正周期和单调递增区间

Answer 1

f(x)=(1-cos2x)/2+(√3/2)sin2x
=(√3/2)sin2x-(1/2)cos2x+1/2
=sin(2x-π/6)+1/2
最小正周期T=2π/2=π
递增区间：
-π/2+2kπ<2x-π/6<π/2+2kπ
-π/3+2kπ<2x<2π/3+2kπ
-π/6+kπ<x<π/3+kπ
所以，递增区间为（-π/6+kπ，π/3+kπ）绝雹k∈Z

祝你开心！希望能帮到你，如族缺果不懂，请追问，祝学习并穗帆进步！O(∩_∩)O