Step 1: Working from the RHS instead:
Let \(\phi=\cos^{-1}(4x^3-3x)\), so \(\cos\phi=4x^3-3x\), and again put \(x=\cos\theta\) with \(\theta\in[0,\pi/3]\).
Step 2: Recognising the triple-angle value:
\(\cos\phi=4\cos^3\theta-3\cos\theta=\cos3\theta\), so \(\phi\) and \(3\theta\) have the same cosine.
Step 3: Matching principal ranges:
Both \(\phi\) (by definition of \(\cos^{-1}\)) and \(3\theta\) (shown above) lie in \([0,\pi]\), and \(\cos\) is one-one there, so \(\phi=3\theta\) exactly, not just up to a coterminal angle.
Final Answer:
So \(\phi=3\theta=3\cos^{-1}x\), proving \(\boxed{3\cos^{-1}x=\cos^{-1}(4x^3-3x)}\).