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(C) Given equation: $\operatorname{Sin}^{-1} x - \operatorname{Cos}^{-1} x = \operatorname{Sin}^{-1}(3x - 2)$.We know that $\operatorname{Sin}^{-1} x

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$5$

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(C) Given equation: $\operatorname{Sin}^{-1} x - \operatorname{Cos}^{-1} x = \operatorname{Sin}^{-1}(3x - 2)$.We know that $\operatorname{Sin}^{-1} x + \operatorname{Cos}^{-1} x = \frac{\pi}{2}$,so $\operatorname{Cos}^{-1} x = \frac{\pi}{2} - \operatorname{Sin}^{-1} x$.Substituting this into the equation: $\operatorname{Sin}^{-1} x - (\frac{\pi}{2} - \operatorname{Sin}^{-1} x) = \operatorname{Sin}^{-1}(3x - 2)$.$2\operatorname{Sin}^{-1} x - \frac{\pi}{2} = \operatorname{Sin}^{-1}(3x - 2)$.Taking $\sin$ on both sides: $\sin(2\operatorname{Sin}^{-1} x - \frac{\pi}{2}) = 3x - 2$.$-\cos(2\operatorname{Sin}^{-1} x) = 3x - 2$.Using $\cos(2	heta) = 1 - 2\sin^2 	heta$,where $	heta = \operatorname{Sin}^{-1} x$,we get $\cos(2\operatorname{Sin}^{-1} x) = 1 - 2x^2$.So,$-(1 - 2x^2) = 3x - 2$.$2x^2 - 1 = 3x - 2 \implies 2x^2 - 3x + 1 = 0$.$(2x - 1)(x - 1) = 0$.Thus,$x = 1$ or $x = \frac{1}{2}$.Check $x = 1$: $\operatorname{Sin}^{-1}(1) - \operatorname{Cos}^{-1}(1) = \frac{\pi}{2} - 0 = \frac{\pi}{2}$. $RHS$: $\operatorname{Sin}^{-1}(3(1) - 2) = \operatorname{Sin}^{-1}(1) = \frac{\pi}{2}$. So $x = 1$ is a solution.Check $x = \frac{1}{2}$: $\operatorname{Sin}^{-1}(\frac{1}{2}) - \operatorname{Cos}^{-1}(\frac{1}{2}) = \frac{\pi}{6} - \frac{\pi}{3} = -\frac{\pi}{6}$. $RHS$: $\operatorname{Sin}^{-1}(3(\frac{1}{2}) - 2) = \operatorname{Sin}^{-1}(-\frac{1}{2}) = -\frac{\pi}{6}$. So $x = \frac{1}{2}$ is a solution.Given $\alpha > \beta$,we have $\alpha = 1$ and $\beta = \frac{1}{2}$.Then $3\alpha + 4\beta = 3(1) + 4(\frac{1}{2}) = 3 + 2 = 5$.

If $\alpha, \beta$ are the solutions of the equation $\operatorname{Sin}^{-1} x - \operatorname{Cos}^{-1} x = \operatorname{Sin}^{-1}(3x - 2)$ and $\alpha > \beta$,then $3\alpha + 4\beta =$

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