Answer
$$A= \frac{23\pi}{18}+2 n\pi,\ \ \text{or}\ \ \frac{35\pi}{18}+2 n\pi$$
where $n $ is integer
Work Step by Step
Given
$$ \sin \left(A-\frac{\pi}{9}\right) =\left( \frac{-1}{2}\right)$$
Since
\begin{align*}
A-\frac{\pi}{9}&=\sin^{-1}\left( \frac{-1}{2}\right)\\
A-\frac{\pi}{9}&=\frac{7\pi}{6},\ \ \text{or}\ \ \frac{11\pi}{6}\\
A&=\frac{23\pi}{18},\ \ \text{or}\ \ \frac{35\pi}{18}
\end{align*}
Since the period of $\cos$ function is $2\pi $ , then the general solution is
$$A= \frac{23\pi}{18}+2 n\pi,\ \ \text{or}\ \ \frac{35\pi}{18}+2 n\pi$$
where $n $ is integer