解.2sin²a-sinacos a-3cos²a=(2sina-3cosa)(sina+cosa)=0∵a∈(0,π/2）∴2sina=3cosa即sina=3/√13,cosa=2/√13sin(a+π/4)=(√2/2)(sina+cosa)=5/√26sin2a=2sinacosa=12/13cos2a+1=2cos²a=8/13所...