$3{\cos ^{ - 1}}x = {\cos ^{ - 1}}(4{x^3} - 3x),\;x \in \left[ {\frac{1}{2},\;1} \right]$
$3{\cos ^{ - 1}}x = {\cos ^{ - 1}}(4{x^3} - 3x),\;x \in \left[ {\frac{1}{2},\;1} \right]$
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Put ${\cos ^{ - 1}}x = \theta .\;\;Then,\;x = \cos \theta$
Now, $\cos 3\theta = )4{\cos ^3}\theta - 3\cos \theta ) = (4{x^3} - 3x)$
$\Rightarrow$ $3\theta = {\cos ^{ - 1}}(4{x^3} - 3x)$
$\Rightarrow$ $3{\cos ^{ - 1}}x = {\cos ^{ - 1}}(4{x^3} - 3x)$
Hence, $3{\cos ^{ - 1}}x = {\cos ^{ - 1}}(4{x^3} - 3x)$
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