Find the number of solutions to the equation $\sin\left(\frac{\pi\sqrt x}4\right)+\cos\left(\frac{\pi\sqrt {2-x}}4\right)=\sqrt 2$ The following question appeared in a JEE Mock exam, held two days ago.
Question:
Find the number of solutions to the equation $\sin\left(\frac{\pi\sqrt x}4\right)+\cos\left(\frac{\pi\sqrt {2-x}}4\right)=\sqrt 2$
My Attempt:
I think if each term is $\frac1{\sqrt 2}$ then it would add upto $\sqrt 2$. So, $x=1$ is a solution.
How can we tell about other solutions, if any?

Let $x = 2 \cos^2 \theta, \theta \in \left[0, \dfrac{\pi}{2}\right]$
Then we have $\sin \left(\dfrac{\pi \cos \theta}{2\sqrt 2}\right) +\cos \left(\dfrac{\pi \sin \theta}{2\sqrt 2}\right) = \sqrt 2 $
Now if $\sin \theta > \cos \theta $ we have $\cos \left(\dfrac{\pi \sin \theta}{2\sqrt 2}\right)<\cos \left(\dfrac{\pi \cos \theta}{2\sqrt 2}\right)$ and hence
LHS  $< \sin \left(\dfrac{\pi \cos \theta}{2\sqrt 2}\right)+\cos \left(\dfrac{\pi \cos \theta}{2\sqrt 2}\right)<\sqrt 2$ (by a well known result) which contradicts equality condition.
Similarly we can argue that $\sin \theta \not < \cos \theta$. Hence we have $\sin \theta = \cos \theta = \dfrac{\pi}{4}$ and at which LHS=RHS.
Thus $\boxed{x=1}$ is the unique solution
