How small must $x$ be to have $\frac{1}{2} \cdot 10^{-8}$ accuracy? 
For small values of $x$, how good is the approximation $cos(x)\approx 1$? How small must $x$ be to have $\frac{1}{2} \cdot 10^{-8}$ accuracy?

My teacher told me, it would be easier to do the second part of the question first, so I need to find out how small $x$ be to have $\frac{1}{2} \cdot 10^{-8}$ accuracy
So as far as I understand, I need to solve for $x$ in the following equation  
$|cos(x)-1|<\frac{1}{2} \cdot 10^{-8} \Rightarrow \\
|x| < \ arccos(1-\frac{1}{2} \cdot 10^{-8}) \Rightarrow \\
x \in \ (-arccos(1-\frac{1}{2} \cdot 10^{-8}),arccos(1-\frac{1}{2} \cdot 10^{-8}))$
but that just seems like a messy answer. Any ideas?

If you haven't learned about Taylor series.
Will you accept that $\sin x \approx x$ when $x$ is small?
and $\sin x < x$ for all $x.$
You can do this geometrically, plot your unit circle and see that for small $x$, the distance traveled about the $x$ axis is very nearly equal to the length of the curve.
$\cos x = \sqrt{1-sin^2 x}$
$\cos x \approx (1-x^2)^{\frac 12}$
In fact, $\cos x$ is slightly greater than (1-x^2)^{\frac 12}$ when $x$ is small, but we will stick with this approximation.
Now do a binomial expansion on $(1-x^2)^{\frac 12}$
$(1-x^2)^{\frac 12} 1 - \frac 12 x^2$
$cos x \approx  1 - \frac 12 x^2 $
Your error is: 
$\frac 12 x^2 < \frac 12 10^{-8}$
If $|x| < 10^{-4}, |\cos x - 1| < \frac 12 10^{-8}$  
