Question

The number of the solutions of the equation $\cos \left(\pi \sqrt{x-4}\right)\cos \left(\pi \sqrt{x}\right)=1$ is

Answer 1

Clearly, $x\ge 4$ (Since $\sqrt{x-4}$ is real) so that $\sqrt{x}$ is alos real.

Again, if $\cos \left(\pi \sqrt{x}\right)<1$ then $\cos \left(\pi \sqrt{x-4}\right)>1$ and if $\cos \left(\pi \sqrt{x}\right)>1$ then $\cos \left(\pi \sqrt{x-4}\right)<1$ ( Since their product $=1$)

But both of these are not possible (Since $\cos \theta$ cannot be greater than $1$)

$\therefore \cos \left(\pi \sqrt{x-4}\right)=1$ and $\cos \left(\pi \sqrt{x}\right)=1$

$\therefore x-4=0$ and $x=0$

But $x=0$ is not possible,

$\therefore x=4$ is the only solution.