มีเเค่ $x=y=0$ หรือป่าวอ่ะครับ
Clearly that $y=0$ then $x=0$ thus if $y\not=0$ Let $t=\sin^2 \pi/10$
the fact $\sin \pi/10 <\dfrac{1}{\sqrt{5}}...1$
Let $\displaystyle\sin \frac{\pi}{10}\ge \frac{1}{5}$
so $\displaystyle \frac{2}{\sqrt{5}}\ge \cos\frac{\pi}{10}=\sin\frac{4\pi}{10}=4\sin\frac{\pi}{10}\cos\frac{\pi}{10}\cos\frac{2\pi}{10}\ge \frac{4}{\sqrt{5}}\cos\frac{\pi}{10}\cos\frac{2\pi}{10}$
Thus, $\displaystyle \frac{1}{2}\ge \cos\frac{\pi}{10}\cos\frac{2\pi}{10}=\frac{1}{2}(\cos \frac{3\pi}{10}+\cos\frac{\pi}{10})>\frac{1}{2}$ contradiction
so equality becomes $(2-t)x^2+(-2y-2yt)x+(1-t)y^2=0$
By the discreminant $x$ will be real if $((-2-2t)y)^2\ge 4(2-t)(1-t)y^2\leftrightarrow t\ge 1/5$
from $y\not =0\rightarrow y^2>0$ so we can devide it both sides
which contradiction so if $y\not =0$ $x$ cannot be real number