The above given equation can be written as $\large\frac{\pi}{2}-\frac{\pi}{6}$$=\sin^{-1}\big(\large\frac{\sqrt 3}{x^2}\big)$
$\Rightarrow \sin\large\frac{\pi}{3}=\frac{\sqrt 3}{x^2}$
$\large\frac{\sqrt 3}{2}=\frac{\sqrt 3}{x^2}$
$x^2=2$
$x=\pm\sqrt 2$
Hence (b) is the correct answer.