${\sin }^{-1} 6x+{\sin }^{-1} 6\sqrt{3}x=-\pi /2\text{ if }x$ is equal to

The given equation can be written as

$\begin{array}{l}{\sin }^{-1} 6x=-\pi /2-{\sin }^{-1} 6\sqrt{3}x\\ \Rightarrow \sin \left({\sin }^{-1} 6x\right)=\sin \left(-\pi /2-{\sin }^{-1} 6\sqrt{3}x\right)\\ \Rightarrow 6x=-\cos \left({\sin }^{-1} 6\sqrt{3}x\right)\\ \Rightarrow 6x=-\sqrt{1-108x^2}\end{array}$

Squaring both sides, we get

$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}36x^2=1-108x^2\Rightarrow 144x^2=1\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x=\pm 1/12\end{array}$

Substituting $x=-1/12$ in the given equation, we get

LHS = 

${\sin }^{-1} \left(-\frac{1}{2}\right)+{\sin }^{-1} \left(-\frac{\sqrt{3}}{2}\right)=-\frac{\pi }{6}-\frac{\pi }{3}=-\frac{\pi }{2}=\text{ RHS }$

Therefore, $x=-1/12$ is a root of the given equation. Again

substituting $x=1/12$ in the given equation, we get LHS = 

$\frac{\pi }{2}$ , so that x = 1/12 is an extraneous root of the given equation. Hence x = –1/12 is the only root.