Let $f:[\frac{1}{2},1]\to [-1,1]$ is given by  $f\left(x\right)=4x^3-3x$ then  $f^{-1}\left(x\right)$ is given by

Let  $x=\cos \theta \Rightarrow \theta ={\cos }^{-1}x$

$f\left(x\right)=4{\cos }^{3}x-3\cos \theta =\cos 3\theta$

$x\in [\frac{1}{2},1]\Rightarrow \cos \theta \in [\frac{1}{2},1]$

$\theta \in [0,\frac{\pi }{3}]$

$3\theta \in [0,\pi ]$

$f\left(x\right)=\cos 3\theta$

$\cos 3\theta =y$

$3\theta ={\cos }^{-1}\left(y\right)$
$\theta =\frac{1}{3}{\cos }^{-1}\left(y\right)$

${\cos }^{-1}x=\frac{1}{3}{\cos }^{-1}\left(y\right)$

$x=\cos \left(\frac{1}{3}{\cos }^{-1}\left(y\right)\right)$

$f^{-1}\left(y\right)=\cos \left(\frac{1}{3}{\cos }^{-1}\left(y\right)\right)$

$f^{-1}\left(x\right)=\cos \left(\frac{1}{3}{\cos }^{-1}\left(x\right)\right)$