Question

solve this question
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Answer 1

$\huge\underline\bold\red{★Answer★}$

${sin}^{ - 1} (1 - x) - 2 {sin}^{ - 1} x = \frac{\pi}{2} \\ = > {sin}^{ - 1} (1 - x) - 2 {sin}^{ - 1} x = {sin}^{ - 1} (1 - x) + {cos}^{ - 1} (1 - x) \\ = > - 2 {sin}^{ - 1} x = {cos}^{ - 1} (1 - x) \\ put \: \: {sin}^{ - 1} x = \alpha \: \: \: \: then \: \: \: \: sin \alpha = x \\ therefore \: \: \: \: \: = > - 2 \alpha = {cos}^{ - 1} (1 - x) \\ = > \cos( - 2 \alpha ) = (1 - x) \\ = > \cos( 2\alpha ) = 1 - x \\ = > 1 - 2 {sin}^{2} \alpha = 1 - x \\ = > 2 {sin}^{2} \alpha = x \\ = > 2 {x}^{2} = x \\ = > 2 {x}^{2} - x = 0 \\ = > x(2x - 1) = 0 \\ \: so \: \: \: x = 0 \: \: \: \: and \: \: \: \: x = \frac{1}{2} \\ note \: \frac{.}{.} \: \: \: x = \frac{1}{2} \: \: doesnot \: satisfy \: the \: equation$