1. Rewrite the equation

Given $\sin^4\alpha + 4\cos^4\beta + 2 = 4\sqrt{2}\sin\alpha\cos\beta,$ with $\alpha,\beta\in[0,\pi]$.

We need to find $\cos(\alpha+\beta)-\cos(\alpha-\beta).$

2. Use a standard identity for the required expression

Recall: $\cos(\alpha+\beta)-\cos(\alpha-\beta)=-2\sin\alpha\sin\beta.$

So it is enough to determine $\sin\alpha\sin\beta$.

3. Transform the given equation into a sum of squares

Let $x=\sin\alpha,\qquad y=\cos\beta.$ Then $x\in[0,1]$ and $y\in[-1,1]$.

The equation becomes $x^4+4y^4+2=4\sqrt2\,xy.$

Now use $x^4+1\ge 2x^2$ and $4y^4+1\ge 4y^2,$ so $x^4+4y^4+2=(x^4+1)+(4y^4+1)\ge 2x^2+4y^2.$

Hence from the given equation, $4\sqrt2\,xy\ge 2x^2+4y^2.$

But by AM-GM / square form, $2x^2+4y^2-4\sqrt2\,xy=(\sqrt2 x-2y)^2\ge 0,$ so $2x^2+4y^2\ge 4\sqrt2\,xy.$

Combining both inequalities gives equality throughout. Therefore, $x^4+1=2x^2,$ $4y^4+1=4y^2,$ $\sqrt2 x=2y.$

4. Solve the equality conditions

From $x^4-2x^2+1=0$ we get $(x^2-1)^2=0 \implies x^2=1.$ Since $x=\sin\alpha\in[0,1]$, we have $x=1 \implies \sin\alpha=1 \implies \alpha=\frac\pi2.$

From $4y^4-4y^2+1=0$ we get $(2y^2-1)^2=0 \implies y^2=\frac12.$ Also from $\sqrt2 x=2y,$ and $x=1$, we get $\sqrt2=2y \implies y=\frac1{\sqrt2}.$ Thus $\cos\beta=\frac1{\sqrt2} \implies \beta=\frac\pi4$ (because $\beta\in[0,\pi]$).

5. Compute the required value

Now