Complete expression for $3\cos^{-1}x$
My Attempt
Let $a=\cos^{-1}x\implies x=\cos a$ $$ \cos3a=4\cos^3a-3\cos a=4x^3-3x=\sin\bigg[\sin^{-1}\Big(4x^3-3x\Big)\bigg]\\ 3a=3\cos^{-1}x=2n\pi\pm\cos^{-1}\Big(4x^3-3x\Big)\\ \boxed{\cos^{-1}\Big(4x^3-3x\Big)=2n\pi\pm\big(3\cos^{-1}x\big)} $$ $$ a=\cos^{-1}x\in [0,\pi]\implies3a\in[0,3\pi] $$ Case 1: $3a\in [0,\pi]$ $$ \color{darkblue}{3\cos^{-1}x=\cos^{-1}\Big(4x^3-3x\Big)}\\ 3\cos^{-1}x\in [0,\pi]\implies\cos^{-1}x\in [0,\pi/3]\implies x\in [1/2,1] $$
Case 2: $3a\in (\pi,2\pi]$ $$ \color{darkblue}{3\cos^{-1}x=2\pi-\cos^{-1}\Big(4x^3-3x\Big)}\\ 3\cos^{-1}x\in (\pi,2\pi]\implies\cos^{-1}x\in (\pi/3,2\pi/3]\implies x\in [-1/2,1/2) $$ Case 3: $3a\in (2\pi,3\pi]$ $$ \color{darkblue}{3\cos^{-1}x=-2\pi+\cos^{-1}\Big(4x^3-3x\Big)}\\ 3\cos^{-1}x\in (2\pi,3\pi]\implies\cos^{-1}x\in (2\pi/3,\pi]\implies x\in [-1,-1/2) $$ How do I confirm that my attempt provides a complete expression for $3\cos^{-1}x$.