anyone have any clever ideas on how to solve this for x?
$\displaystyle cos^2(\pi*16^x)+cos^2(\pi*16^{1-x})-2 = 0$
?
$\displaystyle (cos(\pi*16^x)-1)(cos(\pi*16^x)+1)=-(cos(\pi*16^{1-x})-1)(cos(\pi*16^{1-x})+1)$
?
Thanks for being so nice about me not explaining in the manner you require, my bad. But as to the 'huh', I was firstly confused at how your statement followed from the equation, and secondly confused at why I would/should observe that fact to solve the equation. Is that the only way? Or do you suggest that to be the easiest way? I don't know. But I do know this follows from that observation -
$\displaystyle cos(\pi*16^x) = 1$
and
$\displaystyle cos(\pi*16^x) = -1$
and
$\displaystyle cos(\pi*16^{1-x}) = 1$
and
$\displaystyle cos(\pi*16^{1-x}) = -1$
and since the arccos of both 1 and -1 is $\displaystyle c*\pi$ (or more specifically $\displaystyle c_1*\pi$ and $\displaystyle c_2*\pi$ where $\displaystyle c_1$ is even and $\displaystyle c_2$ is odd) I would now have to find $\displaystyle c_1,c_2,c_3,c_4$ - so unless there is an algebraic method for doing that, splitting the equation in two doesn't help me. Is there anything else you can recommend?
That:
$\displaystyle
\cos^2(\pi*16^x)=1
$
implies that $\displaystyle \pi 16^x=k \pi$ for some integer $\displaystyle k$, which then implies that $\displaystyle 16^x=2^{4x}=k$ but this implies that:
$\displaystyle 4x=\log_2(k)/2$
or:
$\displaystyle x=\log_2(k)/8$.
Similarly (using the other equation): we find that for some integer n:
$\displaystyle 1-x=\log_2(n)/8$.
So:
$\displaystyle \log_2(k)/8=1-\log_2(n)/8$
or:
$\displaystyle \log_2(k/n)=8$
so:
$\displaystyle k/n=2^8.$
so every integer $\displaystyle n$ determines a solution $\displaystyle x=1-\log_2(n)/8$.
(don't rely on the algebra above being right as I cannot check it in detail at present, either I or another helper will correct any mistakes when we get the chance).
RonL