#### Answer

\begin{aligned}
&x_{1} \approx 6.393542\\
&x_{2} \approx 6.439307\\
&x_{3} \approx 6.439117
\end{aligned}

#### Work Step by Step

Given $$f(x)=1-x \sin x, \quad x_{0}=7$$
Since
$$ x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}$$
Then
\begin{aligned}
x_{n+1} &=x_{n}-\frac{1-x_{n} \sin x_{n}}{-\sin x_{n}-x_{n} \cos x_{n}} \\
&=\frac{x_{n} \sin x_{n}+x_{n}^{2} \cos x_{n}+1-x_{n} \sin x_{n}}{\sin x_{n}+x_{n} \cos x_{n}} \\
&=\frac{x_{n}^{2} \cos x_{n}+1}{\sin x_{n}+x_{n} \cos x_{n}}
\end{aligned}
Hence
\begin{aligned}
&x_{1}=\frac{x_{0}^{2} \cos x_{0}+1}{\sin x_{0}+x_{0} \cos x_{0}}= \frac{(7)^{2} \cos (7)+1}{\sin (7)+(7) \cos (7)} \approx 6.393542\\
&x_{2}=\frac{x_{1}^{2} \cos x_{1}+1}{\sin x_{1}+x_{1} \cos x_{1}}= \frac{( 6.393)^{2} \cos ( 6.393)+1}{\sin ( 6.393)+( 6.393) \cos ( 6.393)} \approx 6.439307\\
&x_{3}=\frac{x_{2}^{2} \cos x_{2}+1}{\sin x_{2}+x_{2} \cos x_{2}} = \frac{( 6.439)^{2} \cos ( 6.439)+1}{\sin ( 6.439)+( 6.439) \cos ( 6.439)}\approx 6.439117
\end{aligned}