已知函数fx=2cosx（根号3*sinx+cosx）-1 （1）求f（x）在闭区间0,兀/2上

(1)f(x)=2cosx（√3sinx+cosx）-1
=2√3sinxcosx+2(cosx)^2-1
=√3sin2x+cos2x
=2sin(2x+π/6),
x∈[0,π/2],
u=2x+π/6的值域是[π/6,7π/6],
v=sinu的值域是[-1/2,1],
∴f(x)=2v的最大值=2,最小值=-1.
（2）f(x0)=2sin(2x0+π/6)=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,
∴cos2x0=cos[(2x0+π/6)-π/6]=(-4√3+3)/10.