Answer
$$A= \frac{7\pi}{6} +2 n\pi,\ \ \text{or}\ \ \frac{5\pi}{3}+2 n\pi$$
where $n $ is integer
Work Step by Step
Given
$$ \sin \left(A+\frac{\pi}{12}\right) =\left( \frac{-\sqrt{2}}{2}\right)$$
Since
\begin{align*}
A+\frac{\pi}{12}&=\sin^{-1}\left( \frac{-\sqrt{2}}{2}\right)\\
A+\frac{\pi}{12}&=\frac{5 \pi}{4}\ \ \text{or}\ \ \frac{7\pi}{4}\\
A&=\frac{7\pi}{6},\ \ \text{or}\ \ \frac{5\pi}{3}
\end{align*}
Since the period of $\sin$ function is $2\pi $ , then the general solution is
$$A= \frac{7\pi}{6} +2 n\pi,\ \ \text{or}\ \ \frac{5\pi}{3}+2 n\pi$$
where $n $ is integer