If $\cos^4 \alpha+4\sin^4 \beta-4\sqrt{2}\cos \alpha \sin \beta +2=0$, then find $\alpha$, $\beta$ in $(0,\frac\pi2)$

Question

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

Answer 1

Hint:

For real $\cos^2\alpha,\sin^2\beta$

$$\dfrac{\cos^4\alpha+4\sin^4\beta+1+1}4\ge\sqrt[4]{\cos^4\alpha\cdot4\sin^4\beta}$$

The equality will occur if $\cos^4\alpha=4\sin^4\beta=1$

Answer 2

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.