Using Newton’s method to approximate a point $x$ such that $7 \sin x=x + \cos x$

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I'm comfortable using polynomials for this method, but for sin and cos, would I be using the correct process below? I've pulled this example from an old textbook and was perplexed on how to carry this out.

My work so far

The derivative of $f$, which is

$$f'(x)=\frac{d}{dx}(7 \sin x)=7 \cos x$$

And, applying Newton's method to the below table.

\begin{array}{|c|c|c|c|} \hline x_n& f(x_n) & f'(x_n) & \frac{f(x_n)}{f'(x_n)} & x_n-\frac{f(x_n)}{f'(x_n)} \\ \hline x_1=3 & 0.9878400564 & −6.929947476 & −0.1425465431 & 3.142546543 \\ \hline \end{array}

Should I proceed this way, or should I actually be using $\tan x_n$ to make it simpler? I know that

$$x_{n+1}=x_n-\frac{f(x_n)}{f'x_n)}=x_n-\frac{7\sin x_n}{7\cos x_n}=x_n-\tan x_n$$

Thus

$$x_1=3$$ $$x_2=x_1-\tan x_1=3-\tan 3$$ $$=3.142546543$$

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Let $f(x)= 7 \sin x-x - \cos x$. Then, $f’(x)=7\cos x -1+\sin x$ and the first iteration is

$$x_1=3,\>\>\> f(3)= -1.022, \>\>\>f’(3)=-7.789,\>\>\> x_2=x_1-\frac{f(3)}{f’(3)}= 2.869$$