当X趋向于0时,1-cosx,cos2x,cos3x与ax^n为等价无穷小,求n,a

1-cosxcos2xcos3x=1-cosx(1-2sin^2x)(cosxcos2x-sinxsin2x)
=1-cosx(1-2sin^2x)[cosx(1-2sin^2x)-2sin^2xcosx]
=1-cos^2x(1-2sin^2x)(1-4sin^2x)
=1-(1-sin^2x)(1-2sin^2x)(1-4sin^2x)
=8sin^6x-14sin^4x+7sin^2x,
由于sinx与x为等阶无穷小,而sin^6x和sin^4x相对于sin^2x是高阶无穷小,
因此8sin^6x-14sin^4x+7sin^2x与7x^2为等阶无穷小,即n=2,a=7.