Question

已知函数f(x)=2sin^2(π/4+x)-√3cos2x 1.求f(x)的最小正周期和单调减区间

Answer 1

2sin^2(π/4+x)可以化为1-COS(π/2+2X)
故原式可化为:f(x)=1-COS(π/2+2X)-√3cos2x即:f(x)=1-[COS(π/2+2X)+√3cos2x]
=1-(SIN2X+√3cos2x)
=1-2*SIN(2X+π/3)
故:1.f(x)的最小正周期为:2π/2的绝对值=π
2.要求f(x)单调减区间,需求SIN(2X+π/3)的单调增区间,即令:2Kπ-π/2<=2X+π/3<=2Kπ+π/2
化简得:Kπ-5/12π<=X<=Kπ+π/12
故:f(x)单调减区间为[Kπ-5/12π,Kπ+π/12](K属于R)
应该没算错吧?那边的东东有点忘了,大概方法还是对的.

Answer 2

2sin^2(π/4+x)可以化为1-COS(π/2+2X)
故原式可化为:f(x)=1-COS(π/2+2X)-√3cos2x
即:f(x)=1-[COS(π/2+2X)+√3cos2x
]
=1-(SIN2X+√3cos2x)
=1-2*SIN(2X+π/3)
故:1.f(x)的最小正周期为:2π/2的绝对值=π
2.要求f(x)单调减区间,需求SIN(2X+π/3)的单调增区间,即令:2Kπ-π/2<=2X+π/3<=2Kπ+π/2
化简得:Kπ-5/12π<=X<=Kπ+π/12
故:f(x)单调减区间为[Kπ-5/12π,Kπ+π/12](K属于R)
应该没算错吧?那边的东东有点忘了,大概方法还是对的.

Answer 3

2sin^2(π/4+x)可以化为1-cos(π/2+2x)
故原式可化为:f(x)=1-cos(π/2+2x)-√3cos2x
即:f(x)=1-[cos(π/2+2x)+√3cos2x
]
=1-(sin2x+√3cos2x)
=1-2*sin(2x+π/3)
故:1.f(x)的最小正周期为:2π/2的绝对值=π
2.要求f(x)单调减区间,需求sin(2x+π/3)的单调增区间,即令:2kπ-π/2<=2x+π/3<=2kπ+π/2
化简得:kπ-5/12π<=x<=kπ+π/12
故:f(x)单调减区间为[kπ-5/12π,kπ+π/12](k属于r)
应该没算错吧?那边的东东有点忘了,大概方法还是对的.

Answer 4

2sin^2(π/4+x)可以化为1-COS(π/2+2X)
故原式可化为:f(x)=1-COS(π/2+2X)-√3cos2x
即:f(x)=1-[COS(π/2+2X)+√3cos2x
]
=1-(SIN2X+√3cos2x)
=1-2*SIN(2X+π/3)
故:1.f(x)的
最小正周期
为:2π/2的绝对值=π
2.要求f(x)单调减区间,需求SIN(2X+π/3)的单调增区间,即令:2Kπ-π/2<=2X+π/3<=2Kπ+π/2
化简
得:Kπ-5/12π<=X<=Kπ+π/12
故:f(x)单调减区间为[Kπ-5/12π,Kπ+π/12](K属于R)
应该没算错吧?那边的东东有点忘了,大概方法还是对的.

Answer 5

f(x)=2sin^2(π/4+x)-√3cos2x
=1+sin2x-√3cos2x=1+2sin(2x-π/3)下面就简单了