find all (x;y) real numbers if:

[trfrm]

Ahlan wa Sahlan.

Excuse me ... .

Find all $(x;y)\in \mathbb{R}^{2}$ if :

\[ \log_{2}(\cos^{2}xy+\frac{1}{\cos^{2}xy})=\frac{1}{y^2-2y+2} \]

\[ \cos^2xy+\cos^{-2}xy=2^{(y^2-2y+2)^{-1}}=:w \]

If $z:=\cos^2{xy}$, then $z+z^{-1}=w$ alias $z^2-wz+1=0$ ... .

\[ z=\frac{1}{2}\left(w\pm\sqrt{w^2-4}\right)=\cos^2{xy} \]

\[ w^2=4^{(y^2-2y+2)^{-1}} \]

\[ \frac{1}{2}\left(2^{(y^2-2y+2)^{-1}}\pm\sqrt{4^{(y^2-2y+2)^{-1}}-4}\right)=\cos^2{xy} \]

\[ 2^{(y^2-2y+2)^{-1}-1}\pm\sqrt{4^{(y^2-2y+2)^{-1}-1}-1}=\cos^2{xy} \]

\[ \cos{xy}=\pm\sqrt{2^{(y^2-2y+2)^{-1}-1}\pm\sqrt{4^{(y^2-2y+2)^{-1}-1}-1}} \]

$xy=2n\pi\pm\arccos\left(\pm\sqrt{2^{(y^2-2y+2)^{-1}-1}\pm\sqrt{4^{(y^2-2y+2)^{-1}-1}-1}}\right)$ with $n\in\mathbb{Z}:=\{0,\pm1,\pm2,\pm3,\dots\}$

Thus, $x=y^{-1}\left(2n\pi\pm\arccos\left(\pm\sqrt{2^{(y^2-2y+2)^{-1}-1}\pm\sqrt{4^{(y^2-2y+2)^{-1}-1}-1}}\right)\right)$ ... .

The RHS in last equation depends $y$ only ... .

Note: $(y^2-2y+2)^{-1}-1=\frac{1}{y^2-2y+2}-1=-\frac{y^2-2y+1}{y^2-2y+2}=-\frac{(y-1)^2}{y^2-2y+2}$ ... .

Hosana in excelcis.

[trfrm]

Assalamu'alaikum warahmatullahi wabarakatuh.

$x=y^{-1}\left(2n\pi\pm\arccos\left(\pm\sqrt{2^{(y^2-2y+2)^{-1}-1}\pm\sqrt{4^{(y^2-2y+2)^{-1}-1}-1}}\right)\right)$

If $y=1$, then

$x=1^{-1}\left(2n\pi\pm\arccos\left(\pm\sqrt{2^{(1^2-2\cdot1+2)^{-1}-1}\pm\sqrt{4^{(1^2-2\cdot1+2)^{-1}-1}-1}}\right)\right)$,

$x=2n\pi\pm\arccos\left(\pm\sqrt{2^0\pm\sqrt{4^0-1}}\right)$,

$x=2n\pi\pm\arccos\left(\pm\sqrt{1\pm\sqrt{1-1}}\right)$,

$x=2n\pi\pm\arccos\left(\pm\sqrt{1\pm\sqrt{0}}\right)$,

$x=2n\pi$ or $x=2m\pi\pm\pi\equiv(2m\pm1)\pi$

with $m,n\in\mathbb{Z}:=\{0,\pm1,\pm2,\pm3,\dots\}$ ... .

Terpujilah Kristus.

[trfrm]

Horas.

that you cannot have $y\ne1$?

In general, we have to use table of $\arccos$ and square root or use calculator ... . Obviously, we can find $x$ for some $y$'s ... because $x=f(y)$ ... . However, for a $y$, in general, the solution $x$ is not unique ... .

Gloria in excelsis Deo.