已知函数f(x)=2根号3sinxcosx+2sin^2x-1,x

求最小正周期和单调区间.速求
2024-12-17 15:44:01
推荐回答(3个)
回答1:

f(x)=√3 sin2x+(1-cos2x)-1=√3sin2x-cos2x=2sin(2x-π/6)
最小正周期T=2π/2=π
单调增区间: 2kπ-π/2=<2x-π/6<=2kπ+π/2, 即 kπ-π/6=单调减区间: 2kπ+π/2=<2x-π/6<=2kπ+3π/2,即kπ+π/3=

回答2:

f(x)=√3sin2x-cos2x
=2sin(2x-π/6)
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所以,递增区间为(-π/6+kπ,π/3+kπ)k∈Z

递减区间:
π/2+2kπ<2x-π/6<3π/2+2kπ
2π/3+2kπ<2x<5π/3+2kπ
π/3+kπ所以,递增区间为(π/3+kπ,5π/6+kπ)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

回答3:

f(x)=2根号3sinxcosx+2sin^2x-1=√3sin2x-cos2x=(1/2)sin(2x-π/6)
1)T=2π/2=π
2) 2kπ-π/2≤2x-π/6≤2kπ+π/2 =>kπ-π/6≤x≤kπ+5π/12
=>单调区间增区间为:[kπ-π/6,kπ+5π/12]
同理:单调区间减区间为:[kπ+5π/12,kπ+5π/6]