需要知识:
- ( x n ) ′ = n x n − 1 (x^n)'=nx^{n-1} (xn)′=nxn−1
- ( s i n x ) ′ = c o s x (sinx)'=cosx (sinx)′=cosx
- [ f ( g ( x ) ) ] ′ = f ′ ( g ( x ) ) × g ′ ( x ) [f(g(x))]'=f'(g(x))\times g'(x) [f(g(x))]′=f′(g(x))×g′(x)
推完之后,考虑导函数与x轴的交点。可以设 t = c o s x t=cosx t=cosx,解二元一次方式。然后用 a c o s ( t ) acos(t) acos(t) 反求 x x x