If $3{\sin }^{-1}x=\pi +{\sin }^{-1}(3x-4x^3),$ then the values satisfying $\text{'}x\text{'}$ are

$3{\sin }^{-1}x=\pi +{\sin }^{-1}(3x-4x^3)$

PUT $x=1\Rightarrow 3{\sin }^{-1}\left(1\right)=\pi +{\sin }^{-1}(-1)$

$\begin{array}{l}3\left(\frac{\pi }{2}\right)=\pi -\frac{\pi }{2}\\ \frac{3\pi }{2}\ne \frac{\pi }{2}\\ x=-1\Rightarrow 3{\sin }^{-1}(-1)=\pi +{\sin }^{-1}(-3+4)\end{array}$ 

$\begin{array}{l}3\left(\frac{-\pi }{2}\right)=\pi +{\sin }^{-1}\left(1\right)\\ \frac{-3\pi }{2}=\pi +\frac{\pi }{2}\\ \frac{-3\pi }{2}\ne \frac{3\pi }{2}\end{array}$