the roots & the limit of $2^{x^{\cos(x)}}\sqrt{\cos(x)}=2^{x}$

Question

If $$2^{(2\pi)^{\cos(2\pi)}}\sqrt{\cos(2\pi)}=2^{2\pi}$$

Can you obtain or is it plausible to find the roots and the limit of $$2^{x^{\cos(x)}}\sqrt{\cos(x)}=2^{x}$$

if $0 < \cos(x)$ and $0 < x$ as $x \rightarrow 0$

Answer 1

I'd say it's quite plausible to find the roots! :) Rewrite as $\cos x=4^{x-x^{\cos x}}$.

For $x\ge1$, $x-x^{\cos x}\ge0$ is true because $x^{\cos x}\le x^1$, thus $4^{x-x^{\cos x}}\ge4^0=1$, with equality if and only if $\cos x=1\iff x=2\pi k$ for $k\in\mathbb N$.

For $x<0$, the problem is nonsense. For $x=0$, equality holds.

For $0<x<1$, we have to use $\cos x<1-\frac{x^2}2+\frac{x^4}{24}$, so it gets somewhat messy. Nevertheless, we get that there are no solutions in this interval.

Thus, our solution set is $x=2\pi k$ for $k\in\mathbb N_0$.