The larger $x$ becomes, the worst the approximation of $x=\sin(x)$ becomes. How large can $x$ get before the error is no longer less than $1/(2\times10^{14})$?
What I have tried: I know that the equation for relative error would make this equation $|\sin(x)-x|/|\sin(x)|\le 1/(2\times10^{14})$.
I also tried using Taylor's theorm for but since both x and n are not fixed variables I seem to have 1 equation 2 unknowns. I also tried using this post but unfortunately could not convert it to what I need. estimating the error of $\sin(x) = x$ with Taylor's Theorem
You know the Taylor series converges globally and that it is an alternating series. Therefore $|\sin(x)-x|\leq|x|^3/6$. Now solve $|x|^3/6 \leq 1/(2 \cdot 10^{14})$.
In terms of the Lagrange remainder, the second order Lagrange remainder term is $f^{(2)}(\xi) x^2/2!$ where $\xi$ is between $0$ and $x$. Since you have $|f^{(2)}(\xi)|=|\sin(\xi)|\leq 1$, it is enough to solve $x^2/2 \leq 1/(2 \cdot 10^{14})$. This argument does not rely on the fact that the Taylor series for $\sin$ converges to $\sin$, whereas the first one does.
Because you end up with $|x|<1$, the second argument can be improved by estimating $|\sin(\xi)|$ by $|x|$ instead. In this case you get $|x|^3/2$ on the LHS so the interval where the error is claimed to be small is not that much smaller than in the first argument.