Question

已知函数f（x）=2√3sinxcosx-2sin²x+1

（1）若x∈R，求函数f（x）的单调增区间
（2）求函数f（x）在区间[0，π/2]上的最小值及此时x的值
（3）若f（x0）=6/5，x0∈[π/4，π/2]，求sin2x0的值

Answer 1

f（x）=2√3sinxcosx-2sin²x+1
=√3sin2x+cos2x
=2sin(2x+π/6)
(2) 2kπ-π/2<=2x+π/6<=2kπ+π/2
kπ-π/3<=x<=kπ+π/6
增区间 【kπ-π/3，kπ+π/6】 k∈Z
（2）x∈【0，π/2】
2x+π/6∈【π/6，7π/6】
x=π/2 fmin=-1
(3) f（x0）=6/5
sin(2x0+π/6)=3/5 x0∈[π/4，π/2]，2x0+π/6∈[2π/3，7π/6]，
cos(2x0+π/6)=-4/5
sin2x0=sin[(2x0+π/6)-π/6]
=sin(2x0+π/6)cosπ/6-cos(2x0+π/6)sinπ/6
=3/5*√3/2+4/5*1/2
=(3√3+4)/10