$3{\sin ^{ - 1}}x = {\sin ^{ - 1}}(3x - 4{x^3}),\;x \in \left[ { - \frac{1}{2},\;\frac{1}{2}} \right]$

Put ${\sin ^{ - 1}}x = \theta .\;\;Then,\;\,x = \sin \theta$
Now, $\sin 3\theta = (3\sin \theta - 4{\sin ^3}\theta ) = (3x - 4{x^3})$

$\Rightarrow$ $3\theta = {\sin ^{ - 1}}(3x - 4{x^3})$

$\Rightarrow$ $3{\sin ^{ - 1}}x = {\sin ^{ - 1}}(3x - 4{x^3})$
Hence, $3{\sin ^{ - 1}}x = {\sin ^{ - 1}}(3x - 4{x^3})$