作业帮已知sinα-√3cosα=(2m-3)/(2-m),求m的取值范围
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已知sinα-√3cosα=(2m-3)/(2-m),求m的取值范围
已知sinα-√3cosα=(2m-3)/(2-m),求m的取值范围
已知sinα-√3cosα=(2m-3)/(2-m),求m的取值范围
sinα-√3cosα
=2(1/2sinα-√3/2cosα)
=2sin(a-π/3)=(2m-3)/(2-m)
所以-2≤(2m-3)/(2-m)≤2
解得m≤7/4.
sinα-√3cosα=2(1/2sinα-√3/2cosα)=2(cos60sina-sin60cosa)=2sin(a-60)
-2<=2sin(a-60)<=2 -2<=(2m-3)/(2-m)<=2
sinα-√3cosα=(2m-3)/(2-m)
2*(1/2*sinα-√3/2*cosα)=(2m-3)/(2-m)
2*(sinαcosπ/3-cosαsinπ/3)=(2m-3)/(2-m)
2*sin(α-π/3)=(2m-3)/(2-m)
sin(α-π/3)=(2m-3)/(4-2m)
-1<=sin(α-π/3)<=1
(2m-3)/...
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sinα-√3cosα=(2m-3)/(2-m)
2*(1/2*sinα-√3/2*cosα)=(2m-3)/(2-m)
2*(sinαcosπ/3-cosαsinπ/3)=(2m-3)/(2-m)
2*sin(α-π/3)=(2m-3)/(2-m)
sin(α-π/3)=(2m-3)/(4-2m)
-1<=sin(α-π/3)<=1
(2m-3)/(4-2m)<=1
(2m-3)/(4-2m)-(4-2m)/(4-2m)<=0
[(2m-3)-(4-2m)]/(4-2m)<=0
(2m-3-4+2m)/(4-2m)<=0
(4m-7)/(4-2m)<=0
(4m-7)(4-2m)<=0
(4m-7)(2m-4)>=0
m<7/4或m>=2
因为分母不能为0
所以m<7/4或m>=2
(2m-3)/(4-2m)>=-1
(2m-3)/(4-2m)+(4-2m)/(4-2m)>=0
[(2m-3)+(4-2m)]/(4-2m)>=0
(2m-3+4-2m)/(4-2m)>=0
1/(4-2m)>=0
4-2m>=0
m<=2
因为分母不能为0
所以m<2
综上m<7/4
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