f(x)
=√3sin2x+t+2cos²x
=√3sin2x+cos2x+1+t
=2sin(2x+π/6)+1+t
=2cos(2x-π/3)+1+t
(1)cos(2x-π/3)=1/2且m⊥n
∴f(x)=2+t=0
∴t=-2
(2)t=1,f(x)=2cos(2x-π/3)+2
f(A)=3,则2A-π/3=π/3,∴A=π/3
b=1
S△ABC=½bc*sinA=√3/2
∴c=2
∴a²=b²+c²-2bc*cosA=1+4-2*2*½=3
∴a=√3
谢谢