lf ${sin}^{- 1} x - {cos}^{- 1} x = {sin}^{- 1} (3 x - 2)$ and $x > 0$ then $x =$

0, 1 / 2

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- 1, 1 / 2

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1, - 1 / 2

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1 / 2, 1

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Solution

The correct option is D $1 / 2, 1$
If $s i n^{- 1} x - c o s^{- 1} x = s i n^{- 1} (3 x - 2)$
take sin on both sides,
$s i n s i n^{- 1} x . c o s c o s^{- 1} x - s i n c o s^{- 1} x . c o s s i n^{- 1} x = 3 x - 2$ $[s i n s i n^{- 1} x = x s i n (a - b) = s i n a c o s b - c o s a s i n b]$
$\Rightarrow x^{2} - (\sqrt{1 - x^{2}})^{2} = 3 x - 2$
$\Rightarrow 2 x^{2} - 1 = 3 x - 2$
$\Rightarrow 2 x^{2} - 3 x + 1 = 0$
$\Rightarrow x = 1, \frac{1}{2}$
correct option is D.

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ϵ [- \frac{1}{2}, \frac{1}{2}]

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