f（x）=2√3sinxcosx+2cos²x-1（x∈R）
f(x) = 2√3sinxcosx+2cos²x-1
= √3 * ( 2sinxcosx) + (2cos²x-1)
= 根号3 sin2x + cos2x
= 2 (sin2xcosπ/6+cos2xsinπ/6)
=2 sin(2x+π/6)
最小正周期：2π/2 = π
在区间[0,π/2]
x ∈[0,π/2]
2x ∈[0,π]
2x+π/6 ∈[π/6,π+π/6]
2x+π/6∈[π/6,π/2]时单调增；
2x+π/6∈[π/2,π+π/6]时单调减
2x+π/6=π/2时有最大值,2sinπ/2=2
2x+π/6=π/6时,f(x)=2sinπ/6=1
2x+π/6=π+π/6时,f(x)=2sin(π+π/6)=-1
所以最小值-1,最大值2
f(x0）=6/5
2 sin(2x0+π/6)=6/5
sin(2x0+π/6)=3/5.(1)
x0∈[π/4,π/2]
2x0∈[π/2,π]
2x0+π/6 ∈[2π/3,7π/6]
cos(2x0+π/6) = - 根号{-sin^2(0+π/6)} = - {1-(3/5)^2} = -4/5.(2)
由（1）：sin2x0cosπ/6+cos2x0sinπ/6=3/5
根号3/2 sin2x0 + 1/2 cos2x0 = 3/5
根号3 sin2x0 + cos2x0 = 6/5 .(3)
由（2）：cos2x0cosπ/6 - sin2x0sinπ/6 = -4/5
根号3/2 cos2x0 - 1/2 sin2x0 = -4/5
根号3 cos2x0 - sin2x0 = -8/5 .(4)
(3)+(4)*根号3得：
cos2x0+3cos2x0 = 6/5 -8根号3/5
4cos2x0= 6/5 -8根号3/5
cos2x0= 3/10 - 2根号3 /5
求cos2x0
