Which of the following is the solution set of the equation ${sin}^{- 1} x = {cos}^{- 1} x + {sin}^{- 1} (3 x - 2)$ ?

{\frac{1}{2}, 1}

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[\frac{1}{2}, 1]

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[\frac{1}{3}, 1]

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{\frac{1}{3}, 1}

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Solution

The correct option is A ${\frac{1}{2}, 1}$
${sin}^{- 1} x = {cos}^{- 1} x + {sin}^{- 1} (3 x - 2)$
$\Rightarrow {sin}^{- 1} x = {sin}^{- 1} \sqrt{1 - x^{2}} + {sin}^{- 1} (3 x - 2)$

$\Rightarrow {sin}^{- 1} x = {sin}^{- 1} (\sqrt{1 - x^{2}} \sqrt{1 - {(3 x - 2)}^{2}} + (3 x - 2) x)$
$\Rightarrow x = \sqrt{1 - x^{2}} \sqrt{12 x - 3 - 9 x^{2}} + (3 x^{2} - 2 x)$

$\Rightarrow - 3 x^{2} + 3 x = \sqrt{1 - x^{2}} \sqrt{12 x - 3 - 9 x^{2}}$

$\Rightarrow - 2 x^{3} + 5 x^{2} - 4 x + 1 = 0$

$\Rightarrow - {(x - 1)}^{2} (2 x - 1) = 0$
$\Rightarrow x = 1, \frac{1}{2}$

Hence solution is $x \in {\frac{1}{2}, 1}$

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