From Förberedande kurs i matematik 1

We can translate the equation \displaystyle \sin v=-\frac{1}{2} to the problem of finding those angles in the unit circle which have a y-coordinate of \displaystyle -\frac{1}{2}. If we compare this with the problem that we had in exercise a, where we looked for angles which satisfied \displaystyle \sin v=+\frac{1}{2}, then the situation is the same, except that the angles now lie under, rather than above, the \displaystyle x -axis, due to reflectional symmetry.

Angle \displaystyle 2\pi -\frac{\pi }{6}=\frac{11\pi }{6} Angle \displaystyle \pi +\frac{\pi }{6}=\frac{7\pi }{6}

The two angles which satisfy \displaystyle \sin v=-\frac{1}{2} lie in the third and fourth quadrants and are \displaystyle v=2\pi -\frac{\pi }{6}=\frac{11\pi }{6} and \displaystyle v=\pi +\frac{\pi }{6}=\frac{7\pi }{6}