Let $$(a, b) \subset (0, 2\pi)$$ be the largest interval for which $$\sin^{-1}(\sin\theta) - \cos^{-1}(\sin\theta) > 0$$, $$\theta \in (0, 2\pi)$$, holds. If $$\alpha x^2 + \beta x + \sin^{-1}(x^2 - 6x + 10) + \cos^{-1}(x^2 - 6x + 10) = 0$$ and $$\alpha - \beta = b - a$$, then $$\alpha$$ is equal to:

Part 1: Finding the interval (a, b).

We need $$\sin^{-1}(\sin\theta) - \cos^{-1}(\sin\theta) > 0$$ for $$\theta \in (0, 2\pi)$$.

Let $$t = \sin\theta$$. The condition becomes $$\sin^{-1}(t) > \cos^{-1}(t)$$.

Since $$\sin^{-1}(t) + \cos^{-1}(t) = \frac{\pi}{2}$$, this gives $$\sin^{-1}(t) > \frac{\pi}{4}$$, i.e., $$t > \frac{1}{\sqrt{2}}$$.

So $$\sin\theta > \frac{1}{\sqrt{2}}$$, which gives $$\theta \in \left(\frac{\pi}{4}, \frac{3\pi}{4}\right)$$.

Therefore $$a = \frac{\pi}{4}$$, $$b = \frac{3\pi}{4}$$, and $$b - a = \frac{\pi}{2}$$.

Part 2: Finding $$\alpha$$.

Note that $$x^2 - 6x + 10 = (x-3)^2 + 1 \geq 1$$. For $$\sin^{-1}$$ and $$\cos^{-1}$$ to be defined, the argument must be in $$[-1, 1]$$. So $$x^2 - 6x + 10 = 1$$, giving $$x = 3$$.

Also, $$\sin^{-1}(1) + \cos^{-1}(1) = \frac{\pi}{2} + 0 = \frac{\pi}{2}$$.

At $$x = 3$$: $$9\alpha + 3\beta + \frac{\pi}{2} = 0$$ ... (i)

Given: $$\alpha - \beta = b - a = \frac{\pi}{2}$$ ... (ii)

From (ii): $$\beta = \alpha - \frac{\pi}{2}$$. Substituting in (i):

$$9\alpha + 3\left(\alpha - \frac{\pi}{2}\right) + \frac{\pi}{2} = 0$$

$$12\alpha - \frac{3\pi}{2} + \frac{\pi}{2} = 0$$

$$12\alpha = \pi$$

$$\alpha = \frac{\pi}{12}$$