已知关于x的方程2x*2-【根号3+1】x+m=0的两根sinq,cosq[1]1+sinq+cosq+2sinqcos

x=(√3+1)/4±√{(√3+1)/4]^2-m/2}
由于sin θ^2+cos θ^2=1
1=2((√3+1)/4)^2+2{(√3+1)/4]^2-m/2}=4[(√3+1)/4]^2-m
m=4[(√3+1)/4]^2-1=√3/2
x=(√3+1)/4±√{1/2-[(√3+1)/4]^2}=(√3+1)/4±(√3-1)/4
则x1=√3/2,x2=1/2
即 θ1=30°, θ2=60°
1）当θ=30°,sinθ=0.5,cosθ=√3/2,tanθ=√3/3
sin^2θ/(sinθ-cosθ)+cosθ/(1-tanθ)=0.5*0.5/[0.5-(√3/2)]+(√3/2)/[1-(√3/3)]=(2-√3)/4
1）当θ=60°,sinθ=√3/2,cosθ=0.5,tanθ=√3
sin^2θ/(sinθ-cosθ)+cosθ/(1-tanθ)=(√3/2)*(√3/2)/[(√3/2)-0.5]+0.5/(1-√3)=-(√3+1)/4