若点O为三角形ABC的外心,且2向量OA+2向量OB+3向量OC=0,C为三角形ABC的内角,cos2C=多少

O为三角形ABC的外心=>|OA|=|OB|=|OC|2OA+2OB+3OC=0|2OA+2OB|=|3OC|(2OA+2OB)(2OA+2OB)=9|OC|^24|OA|^2+4|OB|^2+8|OA||OB|cos∠AOB= 9|OC|^28+8cos∠AOB=9cos∠AOB = 1/8similarly|2OA+3OC|=|2OB|=> cos∠AOC= -3/4|2...