Question

已知函数f（x）＝sin2x-2根号3sin^2x-根号3+1,求f(x)的最小正周期及其单调增区间

Answer 1

答：f(x)=sin2x-2√3(sinx)^2-√3+1

=sin2x+√3(1-2(sinx)^2]-2√3+1

=sin2x+√3cos2x-2√3+1

=2*[(1/2)sin2x+(√3/2)cos2x]-2√3+1

=2sin(2x+π/3)-2√3+1

最小正周期T=2π/2=π

单调递增区间满足：

2kπ-π/2