Answer
$$
4e^{-x^{2}}\sin x=x^{2}-x+1
$$
The roots of the given equation, correct to eight decimal places, are 0.21916368 and 1.08422462.
Work Step by Step
$$
4e^{-x^{2}}\sin x=x^{2}-x+1
$$
We first rewrite the equation in standard form:
$$
4e^{-x^{2}}\sin x-x^{2}+x-1=0
$$
Therefore we let
$$
f(x)=4e^{-x^{2}}\sin x-x^{2}+x-1=0
$$
Then
$$
f^{'}(x)=4e^{-x^{2}}[\cos x-2x\sin x]-2x+1
$$
so, Formula 2 (Newton’s method) becomes
$$
\begin{aligned} x_{n+1} &=x_{n}-\frac{f(x)}{f^{'}(x)}\\
\\ &=x_{n}-\frac{ 4e^{-x_{n}^{2}}\sin x_{n}-x_{n}^{2}+x_{n}-1 }{4e^{-x_{n}^{2}}[\cos x_{n}-2x_{n} \sin x_{n}]-2x_{n}+1 }
\end{aligned}
$$
In order to guess a suitable value for $x_1$ we sketch the graphs of $y=4e^{-x^{2}}\sin x $ and $y = x^{2}-x+1 $ in the Figure .
(*) It appears that they intersect at a point whose x-coordinate is somewhat less than 0.2, so let’s take $x_{1} = 0.2 $ as a convenient first approximation. Then Newton’s method gives
$$
\begin{aligned} x_{2} &=x_{1}-\frac{ 4e^{-x_{1}^{2}}\sin x_{1}-x_{1}^{2}+x_{1}-1 }{4e^{-x_{1}^{2}}[\cos x_{1}-2x_{1} \sin x_{1}]-2x_{1}+1 } \\
&=(0.2)-\frac{ 4e^{-(0.2)^{2}}\sin (0.2)-(0.2)^{2}+(0.2)-1 }{4e^{-(0.2)^{2}}[\cos (0.2)-2(0.2) \sin (0.2)]-2(0.2)+1 } \approx 0.21883273
\end{aligned}
$$
repeating we get
$$
x_{3}\approx 0.21916357,
x_{4} \approx 0.21916368 \approx x_{5} ,
$$
(**) It appears that they intersect at a point whose x-coordinate is somewhat less than 1.1, so let’s take $x_{1} = 1.1 $ as a convenient first approximation. Then Newton’s method gives
$$
\begin{aligned} x_{2} &=x_{1}-\frac{ 4e^{-x_{1}^{2}}\sin x_{1}-x_{1}^{2}+x_{1}-1 }{4e^{-x_{1}^{2}}[\cos x_{1}-2x_{1} \sin x_{1}]-2x_{1}+1 } \\
&=(1.1)-\frac{ 4e^{-(1.1)^{2}}\sin(1.1)-(1.1)^{2}+(1.1)-1 }{4e^{-(1.1)^{2}}[\cos (1.1)-2(1.1) \sin (1.1)]-2(1.1)+1 } \approx 1.08432830
\end{aligned}
$$
repeating we get
$$
x_{3}\approx 1.08422462 \approx x_{4}
$$
we conclude that the roots of the given equation, correct to eight decimal places, are 0.21916368 and 1.08422462.
