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Step-by-Step Solution
Step 1: Analyze the given inequality
We need to find the largest interval $(a, b) \subset (0, 2\pi)$ for which
$ \sin^{-1}(\sin \theta) - \cos^{-1}(\sin \theta) > 0, \quad \theta \in (0,2\pi). $
Step 2: Convert $ \cos^{-1}(\sin \theta) $ into an expression involving $ \sin^{-1} $
Recall an important identity for angles where both inverse functions make sense:
$ \cos^{-1}(x) = \frac{\pi}{2} - \sin^{-1}(x). $
Here, let $x = \sin \theta$. Then
$ \cos^{-1}(\sin \theta) = \frac{\pi}{2} - \sin^{-1}(\sin \theta). $
Step 3: Substitute and simplify the inequality
Substitute this into the given inequality:
$ \sin^{-1}(\sin \theta) - \bigl(\tfrac{\pi}{2} - \sin^{-1}(\sin \theta)\bigr) > 0. $
Distribute the negative sign and combine like terms:
$ \sin^{-1}(\sin \theta) - \tfrac{\pi}{2} + \sin^{-1}(\sin \theta) > 0, $
$ 2\,\sin^{-1}(\sin \theta) - \tfrac{\pi}{2} > 0. $
Thus,
$ 2\,\sin^{-1}(\sin \theta) > \frac{\pi}{2}, $
$ \sin^{-1}(\sin \theta) > \frac{\pi}{4}. $
Step 4: Determine the range of $ \theta $
The condition $ \sin^{-1}(\sin \theta) > \tfrac{\pi}{4} $ implies
$ \sin \theta > \sin\bigl(\tfrac{\pi}{4}\bigr) = \tfrac{1}{\sqrt{2}}. $
Hence, the general solution for
$ \sin \theta > \tfrac{1}{\sqrt{2}} $
within $(0, 2\pi)$ corresponds to
$ \theta \in \bigl(\tfrac{\pi}{4}, \tfrac{3\pi}{4}\bigr). $
Therefore, the largest interval $(a, b)$ satisfying the given inequality is
$ a = \tfrac{\pi}{4}, \quad b = \tfrac{3\pi}{4}. $
Step 5: Relate $b-a$ to $\alpha$ and $\beta$
From the interval, we have
$ b - a = \tfrac{3\pi}{4} - \tfrac{\pi}{4} = \tfrac{\pi}{2}. $
The problem states $ \alpha - \beta = b - a $. Therefore,
$ \alpha - \beta = \tfrac{\pi}{2}. $
Step 6: Use the given equation to find $ \alpha $
The provided equation is
$ \alpha x^{2} + \beta x + \sin^{-1}\bigl(x^{2} - 6x + 10\bigr) + \cos^{-1}\bigl(x^{2} - 6x + 10\bigr) = 0. $
Notice that $ x^{2} - 6x + 10 = (x - 3)^{2} + 1. $ A convenient choice is $ x = 3 $, because $ (3 - 3)^{2} + 1 = 1. $
Substitute $ x = 3 $ into the equation:
$ \alpha \cdot 3^{2} + \beta \cdot 3 + \sin^{-1}(1) + \cos^{-1}(1) = 0. $
Recall:
$ \sin^{-1}(1) = \frac{\pi}{2}. $
$ \cos^{-1}(1) = 0. $
Hence, the equation becomes:
$ 9\alpha + 3\beta + \tfrac{\pi}{2} = 0. $
Step 7: Solve for $ \alpha $
We already have $ \alpha - \beta = \tfrac{\pi}{2}. $ From this, express $ \beta = \alpha - \tfrac{\pi}{2}. $
Substitute into $ 9\alpha + 3\beta + \tfrac{\pi}{2} = 0 $:
$ 9\alpha + 3 \Bigl(\alpha - \tfrac{\pi}{2}\Bigr) + \tfrac{\pi}{2} = 0. $
Simplify:
$ 9\alpha + 3\alpha - \tfrac{3\pi}{2} + \tfrac{\pi}{2} = 0, $
$ 12\alpha - \pi = 0, $
$ \alpha = \frac{\pi}{12}. $
Final Answer
$ \displaystyle \frac{\pi}{12}. $
