已知6sinα^2+sinαcosα-2cosα^2=0,α∈[π/2,π)求sin(2α+π)的值

∵ 6sin²α+sinαcosα-2cos²α=0, ∴ 6tan²α+tanα-2=0, ∴ tanα=-2/3或tanα=1/2. ∵ α∈[π/2,π), ∴ tanα=-2/3由万能公式,得sin2α=2tanα/(1+tan²α)=-12/13,cos2α=(1-tan²α)/(1+tan²α)=5/13∴ sin(2α+π/3)=sin2αcos(π/3)+cos2αsin(π/3)=(5√3-12)/26 收起