方程sinx+√3cosx+a=0在区间[0,π/2]内有两个相异的实数根A,B2(sinxcos(π/3)+sin(π/3+x)+a=0sin(x+π/3)=-a/2π/3≤x+π/3≤5π/6-2≤a≤-3^(12/) sin(A+π/3)=sin(π-A-π/3)=sin(π/3-A+π/3)π/3-A=BA+B=π/3...