Step 1: Understand the task.
We add the principal values $\sin^{-1}\left(-\dfrac{1}{2}\right)$ and $\sin^{-1}\left(-\dfrac{\sqrt{3}}{2}\right)$.
Step 2: Recall the principal range.
$\sin^{-1}$ returns angles in $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$, so negative inputs give negative angles.
Step 3: Evaluate the first term.
We need an angle in range whose sine is $-\dfrac{1}{2}$; that is $-\dfrac{\pi}{6}$.
Step 4: Evaluate the second term.
The angle in range with sine $-\dfrac{\sqrt{3}}{2}$ is $-\dfrac{\pi}{3}$.
Step 5: Add them.
$-\dfrac{\pi}{6} - \dfrac{\pi}{3} = -\dfrac{\pi}{6} - \dfrac{2\pi}{6} = -\dfrac{3\pi}{6} = -\dfrac{\pi}{2}$.
Step 6: Sanity check.
The sum $-\dfrac{\pi}{2}$ sits at the edge of the principal range, which is fine since we are merely adding two outputs.
\[ \boxed{-\dfrac{\pi}{2}} \]