∵tanα=3,
∴2sin 2α+3sinα•cosα+5cos 2α
=
2 sin 2 α+3sinα•cosα+5 cos 2 α
sin 2 α+ cos 2 α
=
2 tan 2 α+3tanα+5
tan 2 α+1
=
2× 3 2 +3×3+5
3 2 +1
=
16
5 .
故答案为:
16
5
∵tanα=3,
∴2sin 2α+3sinα•cosα+5cos 2α
=
2 sin 2 α+3sinα•cosα+5 cos 2 α
sin 2 α+ cos 2 α
=
2 tan 2 α+3tanα+5
tan 2 α+1
=
2× 3 2 +3×3+5
3 2 +1
=
16
5 .
故答案为:
16
5