f(x)=cos(2x-π/3)+2sin^2 x=cos2xcosπ/3+sin2xsinπ/3+1-cos2x=1/2cos2x+sin2xsinπ/3+1-cos2x=-1/2cos2x+sin2xsin(π-2π/3)+1=cos2π/3cos2x+sin2xsin2π/3+1=cos(2x-2π/3)+1所以最小正周期是π,对称轴方程是x=...我还想问下，若g(x)=(f(x))^2+f(x),那其值域怎么求？g(x)=(f(x))^2+f(x)=[cos(2x-2π/3)+1]^2+cos(2x-2π/3)+1=[cos(2x-2π/3)]^2+3cos(2x-2π/3)+2=[cos(2x-2π/3)+3/2]^2-1/4因为-1≤cos(2x-2π/3)≤1所以0≤[cos(2x-2π/3)+3/2]^2≤(1+3/2)^2=25/4所以g(x)其值域是[-1/4，6]