首先分解因式看出x=1是解,则（x+1）sinB-xsinC-sinA=0的解亦为x=1,因此有2sinB=sinA+sinC,下面进行化简,sinB=sin(A+C)=2sin(A+C)/2*cos(A+C)/2,sinA+sinC=2sin(A+C)/2*cos(A-C)/2,代入整理得
2cos(A+C)/2=cos(A-C)/2;
2cosA/2*cosC/2-2sinA/2*sinC/2=cosA/2*cosC/2+sinA/2*sinC/2;
cosA/2*cosC/2=3sinA/2*sinC/2;
tanA/2*tanC/2=1/3.