Question

已知函数F(x)=2根号3sinxcosx-2cos^2x+2(1)求F(x)的单调递增区间

Answer 1

F(x)=2根号3sinxcosx-2cos^2x+2
=√3 sin2x-（2cos^2x-1）+1
=√3 sin2x-cos2x+1
=2（√3/2 sin2x-1/2cos2x）+1
=2（sin2x *cosπ/6-sinπ/6*cos2x）+1
=2sin（2x+π/6）+1

F(X)是将sinx的周期缩小一半，在向左平移了π/6个单位，然后在向上平移1个单位

其单调区间和sinx的类似，
增区间：2kπ-π/2<2x-π/6<2kπ+π/2
所以增区间为[kπ-π/6,kπ+π/3]

Answer 2

f(x)=根号3sin2x-cos2x+1
=2sin(2x-π/6)+1
令2kπ-π/2<2x-π/6<2kπ+π/2
kπ-π/6<x<kπ+π/3
所以增区间为[kπ-π/6,kπ+π/3]