If $\cos^4 \alpha+4\sin^4 \beta-4\sqrt{2}\cos \alpha \sin \beta +2=0$, then find $\alpha$, $\beta$ in $(0,\frac\pi2)$ 
If  $\cos^4 \alpha+4\sin^4 \beta-4\sqrt{2}\cos \alpha \sin \beta +2=0$,
where $\displaystyle \alpha, \beta \in \bigg(0,\frac{\pi}{2}\bigg)$. Then value of $\alpha,\beta$ are

Try: I am trying to convert it into sum of square of quantity
like $$(\cos^2 \alpha)^2+(2\sin^2 \beta)^2-2\cdot \cos^2 \alpha \cdot 2\sin^2 \beta-4\sqrt{2}\cos \alpha \sin \beta +2+4\cos^2 \alpha \cdot \sin^2 \beta$$
$$(\cos^2 \alpha--2\sin^2 \beta)^2-4\sqrt{2}\cos \alpha \sin \beta+4\cos^2 \alpha \cdot \sin^2 \beta$$
Now i did not know how to solve it, could some help me

If you just call $x=\cos(\alpha)$ and $y=\sin(\beta)$ you get the 4th degree polynomial equation $x^4-4\sqrt{2}xy + y^4 +2 =0$. I don't see a clever way to solve this directly but this can be solved algebraically. So I would put this equation into mathematica or some similar software and then look a the 4 solutions.
