说明：原题打错了!应该是“已知6sin²a+sinacosa－2cos²a＝0,a∈(π/2,π),求sin(2a+π/2)的值.”
∵6sin²a+sinacosa－2cos²a＝0
==>(3sina+2cosa)(2sina-cosa)=0
又a∈(π/2,π)
∴sina>0,cosa3sina+2cosa=0
==>cosa=-3sina/2
==>sin²a+(-3sina/2)²=1
==>sin²a=4/13
==>cos²a=1-sin²a=9/13
故sin(2a+π/2)=cos(2a)
=cos²a-sin²a
=9/13-4/13
=5/13.