①
f(x)=cos﹙2x－4π／3﹚＋2cos²x
=cos2xcos4π/3+sin2xsin4π/3+1+cos2x
=1/2cos2x-√3/2sin2x+1
=cos(2x+π/3)+1
当2x+π/3=2kπ,k∈Z
即x=kπ-π/6,k∈Z时
f(x)的最大值为1+1=2
此时x的集合为{x|x=kπ-π/6,k∈Z}
②
若f﹙B＋C﹚＝3／2
即cos[2(B+C)+π/3]+1=cos[(2π-2A)+π/3]+1
=cos(2A-π/3)+1=3/2
∴cos(2A-π/3)=1/2
∵ 0