Question

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

• A. $0$
• B. $1$
• C. $2$
• D. $infinite$

Answer 1

Answer: (A)

$0$

we have, $\sqrt{1+\cos 2x}$=$\sqrt{2}{\cos }^{-1}\left(\cos x\right)$
$\Rightarrow \sqrt{1+2 \cos^2 x-1}=\sqrt2 \cos^{-1}(\cos
x)$
$\Rightarrow \sqrt{2\cos }=\sqrt{2}{\cos }^{-1}\left(\cos x\right)$
$\Rightarrow \cos x={\cos }^{-1}\left(\cos x\right)$
$\Rightarrow \cos x=x[\because {\cos }^{-1}(\cos x)=x]$Which is not true for any real value of $x$
Hence, there is no solution possible for the given equation.