6sinx^2+sinxcosx-2cosx^2=0
(2sinx-cosx)(3sinx+2cosx)=0
2sinx=cosx tanx=1/2
3sinx=2cosx tanx=-2/3
sin（2x+π/3)
=sin2xcosπ/3+cos2xsinπ/3
=sinxcosx+√3/2(cos^2x-sin^2x)
=[sinxcosx+√3/2(cos^2x-sin^2x)]/[(sinx)^2+(cosx)^2]
分子,分母同时除以(cosx)^2得：
=[tanx+√3/2-(tanx)^2]/[(tanx)^2+1]
因为tanx=-2/3,所以
=[-2/3+√3/2-(-2/3)^2]/[(-2/3)^2+1]
=(9√3-20)/26
因为tanx=1/2,所以
=[1/2+√3/2-(1/2)^2]/[(1/2)^2+1]
=(2√3+1)/5