The number of real Solutions of the equation $\sqrt {1 + \cos 2x} = \sqrt 2 {\cos ^{ - 1}}(\cos x){\mathop{\rm in}\nolimits} \left[ {\frac{\pi }{2},\pi } \right]$ is

We have, $\sqrt {1 + \cos 2x} = \sqrt 2 {\cos ^{ - 1}}(\cos x),\left[ {\frac{\pi }{2},\pi } \right]$

$\Rightarrow$ $\sqrt {1 + 2{{\cos }^2}x - 1} = \sqrt 2 {\cos ^{ - 1}}(\cos x)$
$\Rightarrow$ $\sqrt 2 \cos x = \sqrt 2 {\cos ^{ - 1}}(\cos x)$

$\Rightarrow$ $\cos x = {\cos ^{ - 1}}(\cos x)$

$\Rightarrow$ $\cos x = x$
which is not true for any real value of $x$.

Hence, there is no Solution possible for the given equation.