Question

已知函数f(x)=2根号3sinxcosx+1-2sin^2x，x属于R

求函数f(x) 的最小正周期和单调递增区间

求详细过程谢谢

Answer 1

f(x)=√3sin2x+cos2x
=2sin)2x+π/6)
所以T=2π/2=π

递增则2kπ-π/2<2x+π/6<2kπ+π/2
kπ-π/3<x<kπ+π/6

所以增区间是（kπ-π/3，kπ+π/6）

Answer 2

f(x)=2√3sinxcosx+1-2(sinx)^2
=√3sin2x+ cos2x
=(1/2)[(√3/2)sin2x+ (1/2)cos2x]
=(1/2)sin(2x+π/6)
小正周期=π
单调递增区间
2kπ- π/2<=2x+π/6 <=2kπ+ π/2
kπ- 2π/3 <=x <=kπ+π/3

Answer 3

f(x)=根号3sin(2x)+cos(2x)
=2sin(2x+π/6)
所以T=2π/w=π，单调增区间-π/2+2kπ<2x+π/6<π/2+2kπ
解得x属于-π/3+kπ，π/6+kπ

Answer 4

2xf(x)=根号3sin2x+cos^2x-sin^2x=根号3sin2x+cos2x=2sin(2x+30度）
最小正周期为π，单调增区间为[kπ-π/3,kπ+π/3]

Answer 5

最小正周期为π，单调递增区间为[kπ,kπ+arcsin(√6/4)/2]∪[kπ+π/2-arcsin(√6/4)/2,kπ+π/2]