Let $\theta =\frac{1}{2}{\sin }^{-1}\left(\frac{2}{3}\right)$, so $\sin (2\theta )=\frac{2}{3}$, and note $\frac{1}{2}{\cos }^{-1}\left(\frac{1}{3}\right)=\frac{\pi }{4}-\theta$ via the identity ${\sin }^{-1}u+{\cos }^{-1}u=\pi /2$ applied carefully. From this, the bracket $\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}\left(\frac{2}{3}\right)$ rearranges so that

$k=\tan \theta +\cot \theta =\frac{1}{\sin \theta \cos \theta }=\frac{2}{\sin (2\theta )}=\frac{2}{2/3}=3$.

So the equation is ${\sin }^{-1}(3x-1)={\sin }^{-1}x-{\cos }^{-1}x$.

Rewrite the RHS using ${\cos }^{-1}x=\frac{\pi }{2}-{\sin }^{-1}x$: RHS $=2{\sin }^{-1}x-\frac{\pi }{2}=-\left(\frac{\pi }{2}-2{\sin }^{-1}x\right)$. Equivalently, ${\sin }^{-1}(3x-1)=\frac{\pi }{2}-2{\cos }^{-1}x\cdot$ (form used to take sine).

Taking sine: $3x-1=\sin \left(\frac{\pi }{2}-2{\cos }^{-1}x\right)=\cos (2{\cos }^{-1}x)=2x^2-1$.

$\therefore 2x^2-3x=0\Rightarrow x=0$ or $x=\frac{3}{2}$.

$x=3/2$ is outside the domain of ${\sin }^{-1}$ (need $|3x-1|\le 1$ and $|x|\le 1$): rejected. $x=0$ satisfies both sides (${\sin }^{-1}(-1)=-\pi /2=0-\pi /2$). Valid.

$\therefore$ Number of solutions $=1$.