#### Answer

\begin{aligned}
x_{1}& \approx 0.2854036\\
x_{2}& \approx 0.24288009\\
x_{3}& \approx 0.24267569
\end{aligned}

#### Work Step by Step

Given $$ f(x)=\cos x-4 x, \quad x_{0}=1 $$
Since
$$ x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}$$
Then
\begin{align*}
x_{n+1}&= x_n -\frac{\cos x_n-4 x_n}{-\sin x_n -4}\\
&=\frac{x_{n} \sin x_{n}+\cos x_{n}}{\sin x_{n}+4}
\end{align*}
Hence
\begin{aligned}
x_{1}&=\frac{x_{0} \sin x_{0}+\cos x_{0}}{\sin x_{0}+4}=\frac{(1) \sin (1)+\cos (1)}{\sin (1)+4} \approx 0.2854036\\
x_{2}&=\frac{x_{1} \sin x_{1}+\cos x_{1}}{\sin x_{1}+4}=\frac{(0.2854) \sin (0.2854)+\cos (0.2854)}{\sin (0.2854)+4} \approx 0.24288009\\
x_{3}&=\frac{x_{1} \sin x_{1}+\cos x_{1}}{\sin x_{1}+4} =\frac{(0.24288) \sin (0.24288)+\cos (0.24288)}{\sin (0.24288)+4} \approx 0.24267569
\end{aligned}