在锐角三角形ABC中,角A,B,C对应的边为a,b,c且B=π/3求2sin^2A+cos（A-C)的取值范围

因为B=π/3,所以A+C=2π/3即C=2π/3 -A则2sin²A+cos（A-C)=1-cos2A +cos(2A-2π/3)=1-cos2A +cos2Acos(2π/3)+sin2Asin(2π/3)=1-(3/2)cos2A+(√3/2)sin2A=1-√3*(√3/2 *cos2A-1/2 *sin2A)=1-√3*cos(2A+π/6)...