Find a complex number $w$ such that $w^2=-sqrt{3} - i$

This is a problem in my undergrad foundations class.
$$\begin{equation} w^2=-\sqrt{3} - i \\w=(-\sqrt{3}-i)^{\frac{1}{2}} \\w=\sqrt{2}\bigg(\cos\frac{5\pi}{12}+i\sin\frac{5\pi}{12}\bigg) \end{equation}$$
So I get to here and the next step is
$$\begin{equation} \sqrt{2}\bigg(\cos\bigg(\frac{5\pi}{12}+\pi\bigg)+i\sin\bigg(\frac{5\pi}{12}+\pi\bigg)\bigg) \end{equation}$$
and this is equal to
$$\begin{equation} w(\cos\pi+i\sin\pi)=-w \end{equation}$$
Can someone help me with understanding the last two steps. Thanks

Answer

$\omega^2 =$$2(-\frac {\sqrt 3}{2} + i \frac 12)\\ 2(\cos \frac {7\pi}{6} + i \sin \frac {7\pi}{6})$

$\omega = \sqrt 2(\cos \frac {7\pi}{12} + i \sin \frac {7\pi}{12})$

And here you are....

When we take the square root of the square of something, there are always two solutions. For example $x^2 = 9 \implies x = \pm 3$

We can put in the $\pm symbol here, but that technique breaks down with higher roots.

A more general approach...

$\omega^2 =2(\cos (\frac {7\pi}{6}+2n\pi) + i \sin (\frac {7\pi}{6}+2n\pi))$

Takes advantage of the period nature of the trig functions.

$\omega =\sqrt 2(\cos (\frac {7\pi}{12}+n\pi) + i \sin (\frac {7\pi}{12}+n\pi))$

And now we can look at just the solutions where the argument is in $[0,2\pi)$

$\omega =\sqrt 2(\cos \frac {7\pi}{12} + i \sin \frac {7\pi}{12}),\sqrt 2(\cos (\frac {7\pi}{12}+\pi) + i \sin (\frac {7\pi}{12}+\pi))$