已知x,y∈(0,π),求满足方程cosX+cosy-cos(x+y)=3/2的x,y的值
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已知x,y∈(0,π),求满足方程cosX+cosy-cos(x+y)=3/2的x,y的值
cosx+cosy=2cos((x+y)/2)*cos((x-y)/2),cos(x+y)=2[cos((x+y)/2)]^2-1
cosX+cosy-cos(x+y)=3/2可以变为2cos((x+y)/2)*cos((x-y)/2)-2[cos((x+y)/2)]^2+1=3/2
整理为4[cos((x+y)/2)]^2-4cos((x+y)/2)*cos((x-y)/2)+1=0①
由△≥0,得16[cos((x-y)/2)]^2-16≥0,即[cos((x-y)/2)]^2≥1,∴要想方程有解必须满足cos((x-y)/2)=1
又x,y∈(0,π),∴(x-y)/2)∈(-π/2,π/2),∴x=y,再将其代入①中,得cosx=1/2
∴x=y=π/3
cosX+cosy-cos(x+y)=3/2可以变为2cos((x+y)/2)*cos((x-y)/2)-2[cos((x+y)/2)]^2+1=3/2
整理为4[cos((x+y)/2)]^2-4cos((x+y)/2)*cos((x-y)/2)+1=0①
由△≥0,得16[cos((x-y)/2)]^2-16≥0,即[cos((x-y)/2)]^2≥1,∴要想方程有解必须满足cos((x-y)/2)=1
又x,y∈(0,π),∴(x-y)/2)∈(-π/2,π/2),∴x=y,再将其代入①中,得cosx=1/2
∴x=y=π/3
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