f(x)=2cos(2x+π/3)+√3(sinx+cosx)^2.若ABC分别为△ABC三个角,且cosB=1/3,f[(π/4)+(C/2)]=√3/2.

f(x)=2cos(2x+π/3)+√3(sinx+cosx)^2=cos2x-√3sin2x+√3(1+sin2x)=cos2x+√3,f[(π/4)+(C/2)]=cos(π/2+C)+√3=-sinC+√3=√3/2,∴sinC=√3/2,cosC=1/2.sinB=(2√2）/3,∴sinA=sin(B+C)=(2√2）/3*1/2+1/3*√3/2=...