f（x）=2根號(hào)3sinxcosx+2cos²x-1
=根號(hào)3sin2x+cos2x
=2[sin2x*cosπ/6+cos2x*sinπ/6]
=2sin(2x+π/6)
(1)T=2π/2=π
最大值2和最小值-2
（2）2sin(2x0+π/6)=6/5
sin(2x0+π/6)=3/5
x0∈閉區(qū)間（π/4,π/2）
2x0+π/6∈閉區(qū)間（2π/3,7π/6）第二、三象限,cos(2x0+π/6)=-4/5
cos2x0=cos[(2x0+π/6)-π/6]
=cos(2x0+π/6)*cosπ/6+sin(2x0+π/6)*sinπ/6
-4/5*根號(hào)3/2+3/5*1/2
=(3-4根號(hào)3)/10