∴2 sinbsinc- 1 = cosa = cos[ 180-(b+ c)]=-cos(b+ c)

∴2sinbsinc+cos(b+c)= 1→2sinbsinc+cosbcosc-sinbsinc= 1

∴ cosb cosc+sinb sinc =1→ cos (b-c) or Cos(C-B)= 1.

So B-C=2kπ(k∈N), while b, C∈(0, 180), ∴B-C=0.

So △ABC is an isosceles triangle.

2, ∫a, b, c into arithmetic progression → 2b = a+c (1).

∫∠b = 30 According to the cosine theorem, CosB=(a? +c？ -B? )/2ac=√3/2 ②

From ①, ② to 3b? = (2+3) communication

S△ABC= 1/2acsinB→ac=6。

∴b=√2(2+√3) = 1，∫sinb sinc = cos 2(a/2)，COSA = 2cos 2 (a/2)- 1。

∴2 sinbsinc- 1 = cosa = cos[ 180-(b+ c)]=-cos(b+ c)

∴2sinbsinc+cos(b+c)= 1→2sinbsinc+cosbcosc-sinbsinc= 1

∴ cosb cosc+sinb sinc =1→ cos (b-c) or Cos(C-B)= 1.

So B-C=2kπ(k∈N), while b, C∈(0, 180), ∴B-C=0.

So △ABC is an isosceles triangle.

2, ∫a, b, c into arithmetic progression → 2b = a+c (1).

∫∠b = 30 According to the cosine theorem, CosB=(a? +c？ -B? )/2ac=√3/2 ②

From ①, ② to 3b? = (2+3) communication

S△ABC= 1/2acsinB→ac=6。

∴b=√2(2+√3) =√( 1+√3)？ =√3 + 1