Answer
$$A= n360^{\circ},\ \ \text{or}\ \ 120^{\circ}+n360^{\circ} $$
where $n $ is integer
Work Step by Step
Given
$$ \sin (A+30^{\circ}) =\left( \frac{1}{2}\right)$$
Since
\begin{align*}
A+30^{\circ}&=\sin^{-1}\left( \frac{1}{2}\right)\\
A+30^{\circ}&=30^{\circ},\ \ \text{or}\ \ 150^{\circ}\\
A&= 0^{\circ},\ \ \text{or}\ \ 120^{\circ}
\end{align*}
Since the period of $\sin$ function is $360^{\circ}$ , then the general solution is
$$A= n360^{\circ},\ \ \text{or}\ \ 120^{\circ}+n360^{\circ} $$
where $n $ is integer