Find the principal value of the equation $\sin \theta = \dfrac{1}{\sqrt{2}}$
$\sin \theta = \dfrac{1}{\sqrt{2}} > 0$
$\therefore \;$ $\theta$ lies in the first or the second quadrants.
Now, $\sin \theta = \dfrac{1}{\sqrt{2}} = \sin \left(\dfrac{\pi}{4}\right)$
$\implies$ $\theta = \dfrac{\pi}{4}$, $\;$ $\pi - \dfrac{\pi}{4} = \dfrac{3 \pi}{4}$
Principal value of $\sin \theta$ is in $\left[- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ $\;$ i.e. $\;$ in first or fourth quadrants.
$\because \;$ $\dfrac{3 \pi}{4} \notin \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$,
$\therefore \;$ Principal value of the given equation is $\dfrac{\pi}{4}$