一道向量题高手来

向量a=（λ+2，λ^2-(cosα)^2),向量b=（m，m/2+sinα),λ,m均为实数，向量a=2向量b，求λ/m的取值范围？

∵向量a=2向量b

∴λ+2=2m,λ²-(cosα)²=2(m/2+sinα)=m+2sinα

∴λ=2m-2, (2m-2)²-(cosα)²=m+2sinα

∴4m²-8m+4-1+(sinα)²=m+2sinα

∴(sinα)²-2sinα+1=-4m²+9m-2,即(sinα-1)²=-4m²+9m-2

∵-1<=sinα<=1,-2<=sinα-1<=0

∴0<=(sinα-1)²<=4, 0<=-4m²+9m-2<=4

4m²-9m+2=(4m-1)(m-2)<=0,4m²-9m+6>=0

∴1/4<=m<=2

λ/m=(2m-2)/m=2-2/m

1/2<=1/m<=4

∴-6<=2-2/m<=1,即λ/m范围为[-6,1]

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