设平面向量a=(√3sin(兀+x),2cosx),b=(-2cosx,cosx),已知函数f(x)=a·b＋m在[0,

向量a=(√3sin(兀+x),2cosx)
=（-√3sinx,2cosx)
b=(-2cosx,cosx),
f(x)=a·b＋m
=2√3sinxcosx+2(cosx)^2+m
=√3sin2x+cos2x+1+m
=2(√3/2sin2x+1/2cos2x)+1+m
=2sin(2x+π/6)+1+m
∵x∈[0,π/2] ∴2x+π/6∈[π/6,7π/6]
∴当2x+π/6=π/2时,f(x)取得最大值3+m
由3+m=6得,m=3
(2)
f(x0)=2sin(2x0+π/6)+4=26/5
得sin(2x0+π/6)=3/5
∵x0∈[兀/4,兀/2].
∴2x0+π/6∈[2π/3,7π/6)
∴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*√3/2+3/5*1/2
=(3-4√3)/10