# How do you change (2sqrt(3),2,-1)   from rectangular to cylindrical coordinates?

First, you have to calculate the radius $r = \sqrt{{x}^{2} + {y}^{2}} = \sqrt{12 + 4} = 4$.
Second, you know that $x = r \setminus \cos \left(\setminus \theta\right)$ and $y = r \setminus \sin \left(\setminus \theta\right)$, therefore, $\setminus \cos \left(\setminus \theta\right) = \setminus \frac{\setminus \sqrt{3}}{2}$ and $y = \setminus \frac{1}{2}$. So, $\setminus \theta = \setminus \frac{\setminus \pi}{6}$ (modulo $2 \setminus \pi$).
Conclusion, the cylindrical coordinates are $\left(4 , \setminus \frac{\setminus \pi}{6} , - 1\right)$.