The number of real solutions of the equation $\sqrt{1 + \cos 2x} = \sqrt{2} \cos^{-1}(\cos x)$ in $\left[ \frac{\pi}{2}, \pi \right]$ is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $0$ ##
We have, $\sqrt{1 + \cos 2x} = \sqrt{2} \cos^{-1}(\cos x)$, $\left[ \frac{\pi}{2}, \pi \right]$
$⇒\sqrt{1 + 2 \cos^2 x - 1} = \sqrt{2} \cos^{-1}(\cos x) \quad [∵\cos 2x = 2 \cos^2 x - 1]$
$⇒ \sqrt{2} \cos x = \sqrt{2} \cos^{-1}(\cos x)$
$⇒\cos x = \cos^{-1}(\cos x)$
$⇒\cos x = x \quad [∵\cos^{-1}(\cos x) = x]$
which is not true for any real value of $x$.
Hence, there is no solution possible for the given equation.