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If $x \in\left(0, \frac{\pi}{2}\right)$ then the value of $\cos ^{-1}\left(\frac{7}{2}(1+\cos 2 x)+\sqrt{\left(\sin ^2 x-48 \cos ^2 x\right) \sin x}\right)$ is equal to
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The correct answer is:
$x-\cos ^{-1}(7 \cos x)$
$y=\cos ^{-1}\left(\frac{7}{2}(1+\cos 2 x)+\sqrt{\left(\sin ^2 x-48 \cos ^2 x\right)} \sin x\right)$
$=\cos ^{-1}\left((7 \cos x)(\cos x)+\sqrt{1-49 \cos ^2 x} \sqrt{1-\cos ^2 x}\right)$
$=\cos ^{-1}(\cos x)-\cos ^{-1}(7 \cos x) \quad[\because \cos x < 7 \cos x]$
$=x-\cos ^{-1}(7 \cos x)$
$=\cos ^{-1}\left((7 \cos x)(\cos x)+\sqrt{1-49 \cos ^2 x} \sqrt{1-\cos ^2 x}\right)$
$=\cos ^{-1}(\cos x)-\cos ^{-1}(7 \cos x) \quad[\because \cos x < 7 \cos x]$
$=x-\cos ^{-1}(7 \cos x)$
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