求函数x-(a+bcosx)sinx为x的5阶无穷小的a,b值

要求$x-(a+bcosx)sinx$为$x$的5阶无穷小，即要求其在$x=0$处的5阶导数为0。\n\n首先求该函数在$x=0$处的导数：\n$f'(x) = 1 - (a+bcosx)cosx - bsinxsinx$ \n$f''(x) = -2bcosxsinx - (a+bcosx)sinx$ \n$f'''(x) = -2bcosxsinx - (a+bcosx)cosx$ \n$f^{(4)}(x) = -3bcosxcosx + 2bcosxsinx + (a+bcosx)sinx$ \n$f^{(5)}(x) = 3bcosxsinx - 3bcosxsinx + 2bcosxcosx - (a+bcosx)cosx$ \n\n将$x=0$代入上式得：\n$f'(0) = 1 - a$ \n$f''(0) = -a$ \n$f'''(0) = -a$ \n$f^{(4)}(0) = a$ \n$f^{(5)}(0) = a$ \n\n由于要求$f^{(5)}(0)=0$，所以$a=0$。\n\n将$a=0$代入$f'(0) = 1 - a$得：\n$1 - a = 1 - 0 = 1$ \n\n所以$a=1$。\n\n综上，$a=1$，$b$可以取任意实数。