Question

急！已知函数fx=2cosx(sinx-√3cosx)+√3,x属于R 求最小正周期和单调

求最小正周期和单调递增区间

Answer 1

f(x)=2sinxcosx-2√3cos²x+√3
=sin2x-2√3(1+cos2x)/2+√3
=sin2x-√3cos2x
=√[1^2+(√3)^2]sin(2x-z)
其中tanz=√3/1=√3
z=π/3
所以f(x)=sin(2x-π/3)
T=2π/2=π

sinx的增区间是(2kπ-π/2,2kπ+π/2)
所以2kπ-π/2<2x-π/3<2kπ+π/2
2kπ-π/6<2x<2kπ+5π/6
kπ-π/12<x<kπ+5π/12
所以单调递增区间(kπ-π/12,kπ+5π/12)

Answer 2

f(x)=2sinxcosx-2√3(cosx)^2+√3
=sin2x-2√3*(cos2x+1)/2+√3
=sin2x+√3cos2x
=2sin(2x+π/3)
所以T=2π/2=π
2x+π/3属于[2kπ-2/π,2kπ+2/π]
x属于[kπ-5π/12,kπ+π/12]时单调递增