Question

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

Answer 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=<x<=kπ+π/3
单调减区间： 2kπ+π/2=<2x-π/6<=2kπ+3π/2,即kπ+π/3=<x<=kπ+5π/6

Answer 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<x<π/3+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<x<5π/6+kπ
所以，递增区间为（π/3+kπ，5π/6+kπ）k∈Z

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Answer 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]