已知函数f(x)=根号3sinwxcoswx+cos^2wx(w>0) 发f(x)最小正周期为π/2

f(x)=(√3/2)sin2wx+(1+cos2wx)/2
=(√3/2)sin2wx+(1/2)cos2wx+1/2
=sin(2wx+π/6)+1/2
最小正周期T=2π/2w=π/2
得：w=2
所以,f(x)=sin(4x+π/6)+1/2
最大值为3/2
递增区间：
-π/2+2kπ<4x+π/6<π/2+2kπ
-2π/3+2kπ<4x<π/3+2kπ
-π/6+kπ/2所以,递增区间为（-π/6+kπ/2,π/12+kπ/2）k∈Z
递减区间：
π/2+2kπ<4x+π/6<3π/2+2kπ
π/3+2kπ<4x<4π/3+2kπ
π/12+kπ/2所以,递增区间为（π/12+kπ/2,π/3+kπ/2）k∈Z

祝你开心!希望能帮到你,如果不懂,请追问,祝学习进步!O(∩_∩)O