Answer
$$
3 \sin x=x, $$
so
$$
f(x)=3 \sin x-x $$
$ \Rightarrow$
$$ f^{\prime}(x)=3 \cos x-1
$$
Equation 2 becomes
$$
\begin{aligned}
x_{n+1}&=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}\\
&=x_{n}-\frac{3 \sin x_{n}-x_{n}}{3 \cos x_{n}-1}
\end{aligned}
$$
From the figure, the positive root of $ f(x)=3 \sin x-x $ is near $2$
So we assume that, $\quad x_{1}=2$
$ \Rightarrow$
$$
\begin{aligned}
x_{2}&=x_{1}-\frac{3 \sin x_{1}-x_{1}}{3 \cos x_{1}-1}\\
&=2-\frac{3 \sin (2)-(2)}{3 \cos (2)-1}\\
&\approx 2.323732
\end{aligned}
$$
and
$$
\begin{aligned}
x_{3}&=x_{2}-\frac{3 \sin x_{2}-x_{2}}{3 \cos x_{2}-1}\\
&=2-\frac{3 \sin (2)-(2)}{3 \cos (2)-1}\\
&\approx 2.279595
\end{aligned}
$$
and so on
$$x_{4} \approx 2.278863 \approx x_{5}$$ .
So the positive root is $2.278863,$ to six decimal places.
Work Step by Step
$$
3 \sin x=x, $$
so
$$
f(x)=3 \sin x-x $$
$ \Rightarrow$
$$ f^{\prime}(x)=3 \cos x-1
$$
Equation 2 becomes
$$
\begin{aligned}
x_{n+1}&=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}\\
&=x_{n}-\frac{3 \sin x_{n}-x_{n}}{3 \cos x_{n}-1}
\end{aligned}
$$
From the figure, the positive root of $ f(x)=3 \sin x-x $ is near $2$
So we assume that, $\quad x_{1}=2$
$ \Rightarrow$
$$
\begin{aligned}
x_{2}&=x_{1}-\frac{3 \sin x_{1}-x_{1}}{3 \cos x_{1}-1}\\
&=2-\frac{3 \sin (2)-(2)}{3 \cos (2)-1}\\
&\approx 2.323732
\end{aligned}
$$
and
$$
\begin{aligned}
x_{3}&=x_{2}-\frac{3 \sin x_{2}-x_{2}}{3 \cos x_{2}-1}\\
&=2-\frac{3 \sin (2)-(2)}{3 \cos (2)-1}\\
&\approx 2.279595
\end{aligned}
$$
and so on
$$x_{4} \approx 2.278863 \approx x_{5}$$ .
So the positive root is $2.278863,$ to six decimal places.