作业帮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](/uploads/image/z/1990914-42-4.jpg?t=y%3Dsinx%2F%282-cosx%29+%3D2sinx%2Acosx%2F%5B3sin%5E2%28x%2F2%29%2Bcos%5E%28x%2F2%29%5D+%3D2%2F%5B3tan%28x%2F2%29%2Bcot%28x%2F2%29%5D+tan%28x%2F2%29%3Dz+--%3Ey%3D2%2F%5B3z%2B1%2Fz%5D%3D2z%2F%5B3z%5E2%2B1%5D+---%3E3z%5E2y-2z%2By%3D0+%E8%A6%81%E4%B8%8A%E5%BC%8F%E6%9C%89%E8%A7%A3%E2%96%B3%3D4-12y%5E2%E2%89%A50+---%3E-%E2%88%9A3%2F3%E2%89%A4y%E2%89%A4%E2%88%9A3%2F3%E5%85%B6%E4%B8%AD%2Cy%3Dsinx%2F%282-cosx%29+%3D2sinx%2Acosx%2F%5B3sin%5E2%28x%2F2%29%2Bcos%5E%28x%2F2%29%5D%E6%98%AF)
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