若sinx+cosx=3分之1,x属于（0,π）,则sin2x+cos2x的值是多少

sinx+cosx=1/3
(sinx+cosx)²=sin²x+cos²x+2sinxcosx=1+sin(2x)=1/9
sin(2x)=-8/9
sinx+cosx=√2sin(x+π/4)=1/3
sin(x+π/4)=(1/3√2)
0π/40<1/3√2<√2/2
3π/4π/2π<2x<3π/2
cos(2x)<0
cos(2x)=-√[1-sin²(2x)]=-√17/9
sin(2x)+cos(2x)=-(8+√17)/9