tan(π+α)=3,α∈(π,3/2π),
所以
tanα=3
所以
sin^2α+3sinαcosα+2cos^2α
=（sin^2α+3sinαcosα+2cos^2α）/(sin²α+cos²α)
=(tan²α+3tanα+2)/(tan²α+1)
=（9+9+2）/(9+1)
=2