1.已知6sin^2α+sinαcosα-2cos^2α=0,α属于[pai/2,pai],求sin(2α+pai/3)的值.
2.已知sinx+cosx=1/√2,求下列个式的值：
（1)sin^3x+cos^3x
(2)sin^4x+cos^4x
(3)tan^2x+cot^2x

1)(sina)^2+sinacosa-2(cosa)^2=0
--->(3sina-cosa)(3sina+2cosa)=0
--->sina=(1/2)cosa或者sina=(-2/3)cosa
--->tana=1/2或者tana=-2/3
pi/2=<a=<pi--->tana<0--->tana=-2/3,
sin2a=2tana/[1+(tana)^2]=2(-2/3)/[1+(-2/3)^2]=-12/13
cos2a=[1-(tana)^2]/[1+(tana)^2]=5/13
sin(2a+pi/3=sin2acos(pi/3)+cos2asin(pi/3)
=(1/2)(-12/13)+(√3/2)(5/13)
=(-12+5√3)/26.
2)sinx+cosx=1/√2
--->1+2sinxcosx=1/2
--->sinxcosx=-1/4
1,(sinx)^3+(cosx)^3
=(sinx+cosx)[(sinx)^2-sinxcosx+(cosx)^2]
=(sinx+cosx)(1-sinxcosx)
=(1/√2)*(1+1/4)
=5/(4√2)
2,(sinx)^4+(cosx)^4
=[(sinx)^2+(cosx)^2]^2-2(sinxcosx)^2
=1-2(-1/4)^2
=5/8
3,(tanx)^2+(cotx)^2
=(sinx/cosx)^2+(cosx/sinx)^2
=[(sinx)^4+(cosx)^4]/(sinxcosx)^2
=(5/8)/(-1/4)^2
=10
=