√3sin2C+2cos^2C+1
=√3sin2C+cos2C+1+1
=2sin(2C+ π/6)+2 =3
sin(2C+ π/6)=1/2
2C+ π/6=5π/6
C=π/3
cosA=(2√2)/3,可得 sinA= 1/3
根据正弦定理
a/sinA=c/sinC
a= c*sinA/sinC= 2/3
2) 2sinA=sinB
即 2a=b
cosC= (a²+b²-c²)/ 2ab= 1/2
解得 a= √3,b=2√3
S= 1/2*absinC= 3√3/2
√3sin2C+2cos^2C+1
=√3sin2C+cos2C+1+1
=2sin(2C+ π/6)+2 =3
sin(2C+ π/6)=1/2
2C+ π/6=5π/6
C=π/3
cosA=(2√2)/3,可得 sinA= 1/3
根据正弦定理
a/sinA=c/sinC
a= c*sinA/sinC= 2/3
2) 2sinA=sinB
即 2a=b
cosC= (a²+b²-c²)/ 2ab= 1/2
解得 a= √3,b=2√3
S= 1/2*absinC= 3√3/2