已知x属于[-π/6,π/3],若方程mcosx-1=cosx+m有解,试求参数m的取值范围
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![已知x属于[-π/6,π/3],若方程mcosx-1=cosx+m有解,试求参数m的取值范围](/uploads/image/z/4943739-3-9.jpg?t=%E5%B7%B2%E7%9F%A5x%E5%B1%9E%E4%BA%8E%5B-%CF%80%2F6%2C%CF%80%2F3%5D%2C%E8%8B%A5%E6%96%B9%E7%A8%8Bmcosx-1%3Dcosx%2Bm%E6%9C%89%E8%A7%A3%2C%E8%AF%95%E6%B1%82%E5%8F%82%E6%95%B0m%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
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已知x属于[-π/6,π/3],若方程mcosx-1=cosx+m有解,试求参数m的取值范围
已知x属于[-π/6,π/3],若方程mcosx-1=cosx+m有解,试求参数m的取值范围
已知x属于[-π/6,π/3],若方程mcosx-1=cosx+m有解,试求参数m的取值范围
若m=1,cosx-1=cosx+1
-1=1,不成立
所以m不等于1
mcosx-1=cosx+m
m不等于1,所以cosx=(m+1)/(m-1)
-π/6<=x<=π/3
所以1/2<=cosx<=1
所以1/2<=(m+1)/(m-1)<=1
1/2<=(m+1)/(m-1)
(m+1)/(m-1)-1/2>=0
(m+3)/[2(m-1)]>=0
(m-1)(m+3)>=0
m<=-3.m>=1
m不等于1
m<=-3,m>1
(m+1)/(m-1)<=1
(m+1)/(m-1)-1<=0
2/(m-1)<=0
m-1<=0
m不等于1
m<1
综上
m<=-3
sin(π/6 @)=sin(π/6)cos@ cos(π/6)sin@令tan(@/2)=xsin@=2x/(1 x^2),cos@=(1-x^2)/(1 x^2)代入 sin( π/6 @)=(5√3-12)/26 并整理,有(5√3 1)x^2-26√3x-25 5√3=0[(5√3 1)x (5-√3)](x-5)=0由π/2<@<π,故x>0,由上式x=5故sin@=2x/(1 x^2)=10/26