Answer
$$
\sin x=x^{2}-2
$$
The roots of the given equation, correct to six decimal places, are -1.061550 and 1.728466.
Work Step by Step
$$
\sin x=x^{2}-2
$$
We first rewrite the equation in standard form:
$$
\sin x-x^{2}+2=0
$$
Therefore we let
$$
f(x)=\sin x-x^{2}+2=0 .
$$
Then
$$
f^{'}(x)=\cos x-2x
$$
so, Formula 2 (Newton’s method) becomes
$$
\begin{aligned} x_{n+1} &=x_{n}-\frac{f(x)}{f^{'}(x)}\\
\\ &=x_{n}-\frac{ \sin x_{n}-x_{n}^{2}+2}{\cos x_{n}-2x_{n}}
\end{aligned}
$$
In order to guess a suitable value for $x_1$ we sketch the graphs of $y=\sin x $ and $y = x^{2}-2 $ in the Figure .
(*) It appears that they intersect at a point whose x-coordinate is somewhat less than -1, so let’s take $x_{1} = -1$ as a convenient first approximation. Then Newton’s method gives
$$
\begin{aligned} x_{2} &=x_{1}-\frac{ \sin x_{1}-x_{1}^{2}+2}{\cos x_{1}-2x_{1}} \\
&=(-1)-\frac{ \sin (-1)-(-1)^{2}+2}{\cos (-1)-2(-1)} \approx -1.062406
\end{aligned}
$$
repeating we get
$$
x_{3}\approx -1.061550 \approx x_{4} ,
$$
(**) It appears that they intersect at a point whose x-coordinate is somewhat less than 2, so let’s take $x_{1} = 2 $ as a convenient first approximation. Then Newton’s method gives
$$
\begin{aligned} x_{2} &=x_{1}-\frac{ \sin x_{1}-x_{1}^{2}+2}{\cos x_{1}-2x_{1}} \\
&=(2)-\frac{ \sin (2)-(2)^{2}+2}{\cos (2)-2(2)} \approx 1.753019
\end{aligned}
$$
repeating we get
$$
x_{3}\approx 1.728710,
x_{4}\approx 1.728466 \approx x_{5},
$$
we conclude that the roots of the given equation, correct to six decimal places, are , -1.061550 and 1.728466.