Inverse function to $\cos(\sqrt{x})$ for $x\in [\pi^2; 4\pi^2]$.

Question

I am having trouble with inverse function to $f(x)=\cos(\sqrt{x})$ for $x\in [\pi^2; 4\pi^2]$.

For $x\in [0; \pi^2]$ it should simply be $\arccos^2(x)$, but I do not completely understand, what to do, if I want to find inverse for $x\in [\pi^2; 4\pi^2]$.

I would guess it would be some $f^{-1}(x)=-\arccos^2 (x) . a + b$? I may be wrong.

Thanks for help.

Answer 1

Solve

$$y=\cos(\sqrt x)$$ for $x\in[\pi^2,4\pi^2]$.

We can write

$$\sqrt x=2k\pi\pm\arccos(y)$$ and look for solutions in $[\pi,2\pi]$. As the arc cosine is in $[0,\pi]$, we have

$$\sqrt x=2\pi-\arccos(y)$$ and finally

$$x=(2\pi-\arccos(y))^2.$$