Question

求y=cosx/(2-sinx)值域

Answer 1

问：求y=cosx/(2-sinx)值域

答：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]