Solving $\sin(\omega t)=- \frac 1 2$ Solve the given trigonometric equation:
   $$\sin(\omega t)=- \frac 1 2$$


Here is my attempt:

$$\sin(\omega t)=- \frac 1 2 = \sin \biggr (\pi +\dfrac{\pi}{6}\biggr )$$

Which yields

$$\omega t = \dfrac{7\pi }{6}$$

or 

$$\sin(\omega t)=- \frac 1 2 = \sin \biggr (2\pi -\dfrac{\pi}{6}\biggr )$$

$$\omega t = \dfrac{11\pi}{6}$$

Is my assumption correct?

Regards!
From the unit circle definition of sine, we see that 

$$\omega t = 2n\pi - \dfrac{\pi}{2} \pm \dfrac{\pi}{3}$$

and we can rephrase that as we wish.
Yes it is right but recall to add the $2k\pi$ term with $k\in \mathbb{Z}$, indeed we have

$$\sin(\omega t)=- \frac 1 2\iff \omega t=\dfrac{7\pi }{6}+2k\pi\,\lor\,\omega t=\dfrac{11\pi }{6}+2k\pi$$