# 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

• $\cos^4 \alpha+4\sin^4\alpha-4\sqrt{2}\cos \alpha \cdot \sin \alpha +2=0 \implies (\cos^2 \alpha-2\sin^2\alpha)^2+4\cos^2\alpha \sin^2 \alpha-4\sqrt{2}\cos \alpha \sin \alpha+2=0\implies (\cos^2\alpha-2\sin^2\alpha)^2+2(\sqrt{2}a\cos \alpha \sin \alpha-1)^2=0$ – DXT Jan 24 at 12:49

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 $$\cos^4\alpha=4\sin^4\beta=1$$
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.