KAMP
Kategori Umum => Matematika => Topik dimulai oleh: trfrm pada Juli 19, 2018, 07:26:25 PM
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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: $(y^2-2y+2)^{-1}-1=\frac{1}{y^2-2y+2}-1=-\frac{y^2-2y+1}{y^2-2y+2}=-\frac{(y-1)^2}{y^2-2y+2}$ ... .
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$x=y^{-1}\left(2n\pi\pm\arccos\left(\pm\sqrt{2^{(y^2-2y+2)^{-1}-1}\pm\sqrt{4^{(y^2-2y+2)^{-1}-1}-1}}\right)\right)$
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 $m,n\in\mathbb{Z}:=\{0,\pm1,\pm2,\pm3,\dots\}$ ... .
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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 ... .