角A=π/3,角B=x,則角C=π-x-π/3=2π/3-x,邊BC=2根號(hào)3,
由正弦定理得BC/sinA=AB/sinC=AC/sinB,
AB=BCsinC/sinA=(2根號(hào)3)sin(2π/3-x)/sin(π/3)
=(2根號(hào)3)sin(x+π/3)/[(根號(hào)3)/2]
=4sin(x+π/3),
AC=BCsinB/sinA=(2根號(hào)3)sinx/sin(π/3)=(2根號(hào)3)sinx/[(根號(hào)3)/2]=4sinx
周長為y=AB+BC+AC=4sin(x+π/3)+2根號(hào)3+4sinx=4[sinxcos(π/3)+cosxsin(π/3)]+2根號(hào)3+4sinx
=4[sinx*(1/2)+cosx*(根號(hào)3)/2]+2根號(hào)3+4sinx=6sinx+2(根號(hào)3)cosx+2根號(hào)3
即f（x）=6sinx+2(根號(hào)3)cosx+2根號(hào)3
B=x>0,且角C=π-x-π/3=2π/3-x>0,所以0