$\sin{(\frac{11\pi}{8}\cos{A})} = \frac{1}{\sqrt{2}}$
$\frac{11\pi}{8}\cos{A} = 2n\pi+\frac{\pi}{4},2n\pi+\frac{3\pi}{4}$
$\cos{A} = \frac{16n+2}{11},\frac{16n+6}{11}$
$A = \cos^{-1}{\frac{16n+2}{11}}, \cos^{-1}{\frac{16n+6}{11}}$
From $-1 \leqslant \cos{A} \leqslant 1$
Case I. $A = \cos^{-1}{\frac{16n+2}{11}}$
$-1\leq \frac{16n+2}{11}\leq 1$
We have only one possible $n$, $n=0$ ---> $A = \cos^{-1}{(\frac{2}{11})}$
Case II. $A = \cos^{-1}{\frac{16n+6}{11}}$
$-1\leq \frac{16n+6}{11}\leq 1$
We have two possible values of $n$,
$n=0$ ---> $A = \cos^{-1}{(\frac{6}{11})}$
$n=-1$ ---> $A = \cos^{-1}{(\frac{-10}{11})}$