How do you solve $\sin x=-\dfrac{1}{2}$?

Hint: We will see in which quadrants we have the sign of sine function as negative. We will look at the definition of an inverse trigonometric function. Using this, we will find the angle for which the value of sine function is $\dfrac{1}{2}$. Then we will get the value for the angle with respect to the quadrant in which it lies such that the value of the sine function is $-\dfrac{1}{2}$.

Complete step by step answer:
We know that all trigonometric ratios are positive in the first quadrant. Then, in the second quadrant, the sine function is positive. In the third quadrant, the value of tan is positive. And in the fourth quadrant, the cosine function is positive.
This means that the sine function is negative in the third quadrant and the fourth quadrant. So, the value of $x$ must lie between $\pi $ and $2\pi $.
The inverse trigonometric function is the inverse function of trigonometric function. Using this function, we can obtain the value of the angle using the given trigonometric ratios. So, we have the following,
${{\sin }^{-1}}\left( \dfrac{1}{2} \right)=x$
We know that $\sin \dfrac{\pi }{6}=\dfrac{1}{2}$
Since the angle can be in the third quadrant, we have the value of the angle as,
$\begin{align}
  & x=\pi +\dfrac{\pi }{6} \\
 & \therefore x=\dfrac{7\pi }{6} \\
\end{align}$
Also, the angle can be in the fourth quadrant. Therefore, we have
$\begin{align}
  & x=2\pi -\dfrac{\pi }{6} \\
 & \therefore x=\dfrac{11\pi }{6} \\
\end{align}$
Therefore, the possible values for the angle such that its sine function gives the value as $-\dfrac{1}{2}$ are
$x=\dfrac{7\pi }{6}\text{ or }x=\dfrac{11\pi }{6}$

Note:
It is important to understand the properties of trigonometric functions in different quadrants. We have different identities and rules that help us in computing the value of a trigonometric function or the value of an angle. It is essential that we know the values of trigonometric functions for standard angles for such type questions.