= = = & gtcosB-(2cos？ B- 1)=0

= = = & gtcosB-2cos？ B+ 1=0

= = = & gt2cos？ B-cosB- 1=0

= = = & gt(2cosB+ 1)(cosB- 1)=0

= = = & gtCosB=- 1/2 or cosB= 1.

= = = & gtB=2π/3, or B=0 (rounded)

So, B=2π/3.

It is known that A=π/4

Therefore, C=π-(2π/3)-(π/4)=π/ 12.

From the previous knowledge, B=2π/3.

So, cosB=(a? +c？ -B? )/(2ac)=- 1/2

= = => Answer? +c？ -B? =-communication

Known: a? +c？ =b-ac+2

= = = & gtb-ac+2-b？ =-communication

= = = & gtb？ -b-2=0

= = = & gt(b-2)(b+ 1)=0

= = = & gtB=2, or b =- 1 (< 0, omitted).

Sine theorem is: a/sinA=b/sinB.

= = = & gta/(√2/2)=2/(√3/2)

= = = & gta=(2/3)√6

Therefore, △ ABC area = (1/2) ABS Inc.

=( 1/2)*((2/3)√6 * 2 * sin(π/ 12)

=(2√6/3)*sin[(π/3)-(π/4)]

=(2√6/3)*[(√3/2)*(√2/2)-( 1/2)*(√2/2)]

=(2√6/3)*[(√6-√2)/4]

=(3-√3)/3.