(1) f(x)=2√3sinxcosx+2cos^2-t.
=√3sin2x+1+cos2x-t.
=2[(√3/2sin2x+(1/2)cos2x]+1-t.
=2sin(2x+π/6)+1-t.
令 f(x)=0,则,2sin(2x+π/6)+1-t=0.
由x∈[0,π/2],得π/6≤2x+π/6≤7π/6,所以,-1/2≤sin(2x+π/6)≤1,
故0≤2sin(2x+π/6)+1≤3
∵t∈[0,3].
(2) f(x)=2sin(2x+π/6)+1-t.
若t=3,f(A)=-1,
f(A)=2sin(2A+π/6)+1-3=-1.
2sin2A+π/6=1.
sin(2A+π6)=1/2.
2A+π/6=π/6,--->A=0 不合题设要求,舍去.
或2A+π/6=5π/6.
2A=5/6-π/6=2π/3.
∴ A=π/3.
∵b+c=2.
由三角形的边的关系:两边之和大于第三边,即 b+c>a,∴a