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

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