# Question #57bde

Feb 17, 2017

$\frac{7 \pi}{6} + 2 k \pi$
$\frac{11 \pi}{6} + 2 k \pi$

#### Explanation:

${\sin}^{-} 2 \left(\frac{1}{2}\right) - \to \arcsin \left(- \frac{1}{2}\right)$
Trig table of special arcs and unit circle give:
$\sin x = - \frac{1}{2}$ --> arc $x = - \frac{\pi}{6} + 2 k \pi$ and
arc $x = \pi + \frac{\pi}{6} = \frac{7 \pi}{6} + 2 k \pi$
Because the arc $\frac{\pi}{6}$ is co-terminal to $\frac{11 \pi}{6}$, then, the answers are:
$x = \frac{7 \pi}{6} + 2 k \pi$
$x = \frac{11 \pi}{6} + 2 k \pi$