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$3 \cos ^{-1} x=\cos ^{-1}\left(4 x^3-3 x\right), x \in\left[\frac{1}{2}, 1\right]$
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Verified Answer
Let $\cos ^{-1} x=\theta$
$$
x=\cos \theta
$$
R.H.S $=\cos ^{-1}\left(4 x^3-3 \cos x\right)$
$=\cos ^{-1}\left(4 \cos ^3 \theta-3 \cos \theta\right)$
$=\cos ^{-1}(\cos 3 \theta)\left[\because \cos 3 \theta=4 \cos ^3 \theta-3 \cos \theta\right]$ $=3 \theta=3 \cos ^{-1} x=$ L.H.S.
$$
x=\cos \theta
$$
R.H.S $=\cos ^{-1}\left(4 x^3-3 \cos x\right)$
$=\cos ^{-1}\left(4 \cos ^3 \theta-3 \cos \theta\right)$
$=\cos ^{-1}(\cos 3 \theta)\left[\because \cos 3 \theta=4 \cos ^3 \theta-3 \cos \theta\right]$ $=3 \theta=3 \cos ^{-1} x=$ L.H.S.
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