y=sinx/(2-cosx) =2sinx*cosx/[3sin^2(x/2)+cos^(x/2)] =2/[3tan(x/2)+cot(x/2)] tan(x/2)=z -->y=2/[3z+1/z]=2z/[3z^2+1] --->3z^2y-2z+y=0 要上式有解△=4-12y^2≥0 --->-√3/3≤y≤√3/3其中,y=sinx/(2-cosx) =2sinx*cosx/[3sin^2(x/2)+cos^(x/2)]是
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y=sinx/(2-cosx) =2sinx*cosx/[3sin^2(x/2)+cos^(x/2)] =2/[3tan(x/2)+cot(x/2)] tan(x/2)=z -->y=2/[3z+1/z]=2z/[3z^2+1] --->3z^2y-2z+y=0 要上式有解△=4-12y^2≥0 --->-√3/3≤y≤√3/3其中,y=sinx/(2-cosx) =2sinx*cosx/[3sin^2(x/2)+cos^(x/2)]是
y=sinx/(2-cosx)
=2sinx*cosx/[3sin^2(x/2)+cos^(x/2)]
=2/[3tan(x/2)+cot(x/2)]
tan(x/2)=z
-->y=2/[3z+1/z]=2z/[3z^2+1]
--->3z^2y-2z+y=0
要上式有解△=4-12y^2≥0
--->-√3/3≤y≤√3/3
其中,y=sinx/(2-cosx)
=2sinx*cosx/[3sin^2(x/2)+cos^(x/2)]是如何变化出来的?
y=sinx/(2-cosx) =2sinx*cosx/[3sin^2(x/2)+cos^(x/2)] =2/[3tan(x/2)+cot(x/2)] tan(x/2)=z -->y=2/[3z+1/z]=2z/[3z^2+1] --->3z^2y-2z+y=0 要上式有解△=4-12y^2≥0 --->-√3/3≤y≤√3/3其中,y=sinx/(2-cosx) =2sinx*cosx/[3sin^2(x/2)+cos^(x/2)]是
是分子和分母同时乘以2sinxcosx得出来的
你题目都写错了
y=sinx/(2-cosx)
=2sin(x/2)*cos(x/2)/[3sin^2(x/2)+cos^2(x/2)]
=2/[3tan(x/2)+cot(x/2)]
二倍角公式，自己去看去