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

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