∴cosB= 1/2

And \b are angles in △ABC.

∴B∈(0,π)

∴B=π/3

2、∫cos(A-C)= 0

∴A-C=π/2 - ①

At △ABC

A+C=π-B=2π/3 - ②

∴A=7π/ 12

C=π/ 12

3、sinA = sin 7π/ 12 = sin(π/3+π/4)=(√6+√2)/4

sinC = sinπ/ 12 = sin(π/3-π/4)=(√6-√2)/4

4、S△ABC= 1/2acsinB

=√3/4ac

ac=4

log 4 Sina+log 4 sinc = log 4(Sina sinc)=- 1

∴sinAsinC= 1/4

5. Sine theorem: a/sinA=c/sinC

asinC=csinA= 1

Replace Sina and sinC to get

a=√6+√2

c=√6-√2

Substitute a? +c？ -B? = communication

b=2√3