题目

已知函数fx=2sin（x-派/3）cosx+sinxcosx+根号（3）sin^2 x
1 求fx单调区间

提问时间：2020-11-03

答案

答：
f(x)=2sin(x-π/3)cosx+sinxcosx+√3(sinx)^2
=sin(x-π/3+x)+sin(x-π/3-x)+sinxcosx+(√3/2)(1-cos2x)
=sin(2x-π/3)-√3/2+(1/2)sin2x+√3/2-(√3/2)cos2x
=sin(2x-π/3)+sin(2x-π/3)
=2sin(2x-π/3)
所以：
单调递增区间满足：2kπ-π/2