Finding for what $x$ values the error of $\sin x\approx x-\frac {x^3} 6$ is smaller than $10^{-5}$ 
Find for what $x$ values the error of $\sin x\approx x-\frac {x^3} 6$ is smaller than $10^{-5}$

I thought of two ways but got kinda stuck:


*

*Since we know that $R(x)=f(x)-P(x)$ then we could solve: $\sin x-x+\frac {x^3} 6<10^{-2}$ but I have no idea how to do it, deriving the expression to a more simple expression makes it loose the information on $10^{-5}$.

*Find for what $x,c$ we have $R_5(x)=\frac {\cos (c)x^5}{5!}<10^{-5}$ but can I simply choose $c=0$? then the answer would be: $x<\frac {(5!)^{1/5}}{10}$ which looks like it's right from the graph of $R(x)$ in 1. But why can I choose $c$? what if it was a less convenient function?
Note: no integrals.

Consider
$$
\sin(x) = x - \frac{x^3}{6} + \frac{x^5}{120} - \ldots.
$$
This series is alternating, and the terms are strictly decreasing in magnitude if $|x| < 1$. So we get 
\begin{align*}
\text{estimate} - \text{ reality} &= x - \frac{x^3}{6} - \left(x - \frac{x^3}{6} + \frac{x^5}{120} - \ldots\right) \\
&=\frac{x^5}{120} - \ldots \\
&\leq \frac{x^5}{120} 
\end{align*}
The last estimate works only because the terms are strictly decreasing in magnitude. So it suffices to take $x$ satisfying
$$
-10^{-5} < \frac{x^5}{120} < 10^{-5}
$$
But this is not a necessary condition. 
