Hello しおり,
The first thing I would do is plot:
$$\sin(x)=x^2-3x+1$$
in order to get an idea where the roots are:
View attachment 1670
So, we see the smaller root is about $0.25$ and the larger root is about $2.75$.
Now, as you did, I would define:
$$f(x)=x^2-3x+1-\sin(x)$$
Hence:
$$f'(x)=2x-3-\cos(x)$$
Newton's method gives us the recursive algorithm:
$$x_{n+1}=x_n-\frac{f\left(x_n \right)}{f'\left(x_n \right)}$$
Using our function $f(x)$, we have:
$$x_{n+1}=x_n-\frac{x_n^2-3x_n+1-\sin\left(x_n \right)}{2x_n-3-\cos\left(x_n \right)}=\frac{2x_n^2-3x_n-x_n\cos\left(x_n \right)-\left(x_n^2-3x_n+1-\sin\left(x_n \right) \right)}{2x_n-3-\cos\left(x_n \right)}$$
$$x_{n+1}=\frac{x_n^2-x_n\cos\left(x_n \right)+\sin\left(x_n \right)-1}{2x_n-3-\cos\left(x_n \right)}$$
i) The smaller root: $$x_0=0.25$$
$$x_1\approx0.253025027391$$
$$x_2\approx0.268799448273$$
$$x_3\approx0.268881342724$$
$$x_4\approx0.268881344942$$
$$x_5\approx0.268881344942$$
ii) The larger root: $$x_0=2.75$$
$$x_1\approx2.77019710423$$
$$x_2\approx2.77005756932$$
$$x_3\approx2.77005756269$$
$$x_4\approx2.77005756269$$
Hence, the two roots of the given equation, rounded to 6 decimal places, are:
$$x\approx 0.268881,\,2.770058$$
