You should always draw the picture when you try to find the modulus and argument.
In an Argand Diagram, the complex number $x+\mathrm i y$ corresponds to the point $(x,y)$ on the $xy$-plane.
Put the point $(\sqrt 3,-1)$ on the $xy$-plane, and join it to the origin by a chord (part of a line).
You can see the chord joining the origin to the complex number as the hypotenuse of a right-angled triangle. The length of that chord is the modulus $|z|$, which can be found using Pythagoras:
$$|z|^2 = \left(\sqrt 3\right)^2 + 1^2 = 4$$
This gives $|z| = 2$, since $|z| \ge 0$.
The argument is the angle of the chord measures from "3 o'clock", i.e. the positive real axis. Anti-clockwise rotations are measured as positive, while clockwise rotations are measure as negative.
This may seem odd, but right is positive and left is negative on the number line.
We can use basic SohCahToh Trigonometry to find the argument. First, find the angle between the chord and the $y$-axis, then subtract from $90^{\circ}$ or $\frac{1}{2}\pi$ rads to get the angle between the chord and the positive $x$-axis.
$$\tan \theta = \frac{\mbox{opp}}{\mbox{adj}} = \frac{\sqrt 3}{1} = \sqrt 3$$
The angle between the chord and the $y$-axis is then $\arctan \sqrt 3 = 60^{\circ}$ or $\frac{1}{3}\pi$ rads. That means the argument, i.e. the angle between the chord and the positive $x$-axis (remembering that clockwise is negative) is $-\frac{1}{6}\pi$ or $-30^{\circ}$.
Finally, $|z|=2$ and $\arg z = -\frac{1}{6}\pi$, so
$$z = r(\cos \theta + \mathrm i \sin \theta) = 2\left(\cos\left(-\frac{1}{6}\pi\right)+\mathrm i \sin\left(-\frac{1}{6}\pi\right)\right)$$
Best Answer
There is no such an angle except using the inverse trigonometri functions.
Otherwise, you can approximate it since it looks to be quite close to $\frac \pi 4$ and using a truncated Taylor expansion around this value, you would get $$\tan^{-1}(x)=1+2 \left(x-\frac{\pi }{4}\right)+O\left(\left(x-\frac{\pi }{4}\right)^2\right)$$ and then, ignoring the higher order terms, you could solve $$1+2 \left(x-\frac{\pi }{4}\right)=\frac{\sqrt{3}}2\implies x=\frac{\sqrt{3}+\pi-2 }{4}\approx 0.7184 $$ while the exact value would be $0.7137$.
You could also use very nice approximations for $\sin(x)$ or $\cos(x)$ (have alook here).
Edit
Sooner or later, you will learn that, better than with Taylor series, functions can be locally approximated using Padé approximants. Using the simplest around $x=a$, we have
$$\tan(x)=\frac{1+(x-a)}{1-(x-a)}$$ Using it for your problem, we just need to solve $$\frac{1+(x-\frac \pi 4)}{1-(x-\frac \pi 4)}=\frac{\sqrt{3}}2\implies x=\frac \pi 4+4 \sqrt{3}-7\approx 0.7136$$ which is much better.