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求y=cosx/(2-sinx)值域
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咨询记录 · 回答于2021-06-28
求y=cosx/(2-sinx)值域
y=cosx/(2-sinx)定义域x为任意实数整理得:cosx=2y-ysinxysinx+cosx=2y√(y^2+1) * sin(x+p)=2ysin(x+p)=2y/√(y^2+1)∈[-1,1]所以:-√(y^2+1)<=2y<=√(y^2+1)所以:|2y|<=√(y^2+1)两边平方:4y^2<=y^2+13y^2<=1y^2<=1/3-√3/3<=y<=√3/3值域为[-√3/3,√3/3]