Question

已知函数f（X)=sin²x+2根号3sinxcosx-cos²x

（1） 求函数f（x）的单调增区间 （2） 求函数的最大值和取最大值时x的集合

Answer 1

f(x)=sin²x+2√3sinxcosx-cos²x

=2√3sinxcosx-(cos²x-sin²x)
=√3sin(2x)-cos(2x)
=2sin(2x-π/6)
1)2kπ-π/2≤2x-π/6<2kπ+π/2
kπ-π/6≤x<kπ+π/3
2kπ+π/2≤2x-π/6<2kπ+3π/2
kπ+π/3≤x<kπ+5π/6
单增区间：[kπ-π/6,kπ+π/3]
单减区间：[kπ+π/3,kπ+5π/6]
2)f(x)最大值为2
sin(2x-π/6)=1

2x-π/6=2kπ+π/2

x=kπ+π/3

Answer 2

f(x)=sin²x+2√3sinxcosx-cos²x
=sin²x-cos²x+2√3sinxcosx
=-cos2x+√3sin2x
=2(√3/2sin2x-1/2cos2x)
= -2(sin2xcos2π/3+sin2π/3cos2x)
= -2sin(2x+2π/3)
单调减区间： 2kπ-π/2<2x+2π/3<2kπ+π/2
2kπ-7π/6<2x<2kπ-π/6
kπ-7π/12<x<kπ-π/12
单调增区间： 2kπ+π/2<2x+2π/3<2kπ+3π/2
2kπ-π/6<2x<2kπ+5π/6
kπ-π/12<x<kπ-5π/12
当 2x+2π/3=3π/2时，2x=5π/6
即 x=5π/12 f(x)最大值=2

Answer 3

cos(a+b)=cosacosb-sinasinb

f（X)=sin²x+2√3sinxcosx-cos²x=2[(√3/2)sin2x-(1/2)cos2x=-2cos(2x+π/3)

这样会了吧？

Answer 4