Question

f(x)=-2sin^2x+2根号3sinxcosx+1，1\求f(x)的最小正周期及对称轴

2\ 在【-π/2,π/2】范围内求f(x)的单调减区间

Answer 1

f(x)=-2sin²x+2√3sinxcosx+1
=cos2x-1+√3sin2x+1
=√3sin2x+cos2x
=2sin(2x+∏/6)
1)T=2∏/2=∏
对称轴2x+∏/6=∏/2+k∏
x=∏/6+k∏/2
2) -π/2<x<π/2
-π<2x<π
-5π/6<2x+π/6<7π/6
单调减区间(π/2,7π/6)

Answer 2

f(x)=-2sin^2x+2根号3sinxcosx+1
=(1-2sin^2x)+(根号3)*(2sinxcosx)
=cos2x+(根号3)*sin2x
=2sin(2x+(π/6))
最小正周期T=2π/2=π

在【-π/2,π/2】范围内,
当f(x)=2, 2x+(π/6)=π/2, x=π/6
当f(x)=-2, 2x+(π/6)=-π/2, x=-π/3
所以，在【-π/2,-π/3】区间，f(x)单调递减
在【π/6,π/2】区间，f(x)单调递减