sin2C+√3*cos(A+B)
=sin2C+√3cos(π-C)
=sin2C-√3cosC
=2sinC*cosC-√3cosC
=cosC(2sinC-√3)=0
所以cosC=0或者2sinC-√3,求得C=π/2,π/3,2π/3
因为a>c,所以角A>角C,所以C<π/2,所以C=π/3
cosC=(a^2+b^2-c^2)/2ab=(16+b^2-13)/8b=1/2
求得b=1,3,所以三角形面积=absinC/2=√3,3√3