Question

...pls help..............

Answer 1

Given: $\large\rm{ \sin^{-1} (1-x)-2 \ \sin^{-1} \ x = \dfrac{\pi}{2}}$

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To find: value of x

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Answer: consider x as sin y

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$\large\rm{ \therefore \sin^{-1} (1- \sin \ y) - 2y = \dfrac{\pi}{2}}$

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$\large\rm{ \implies \sin^{-1} (1- \sin \ y) = \dfrac{\pi}{2}+ 2y}$

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$\large\rm{ \implies 1 - \sin \ y = \sin \Bigg ( \dfrac{\pi}{2} + 2y \Bigg ) }$

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$\large\rm{ 1 - \sin \ y = 2 \ \cos \ y}$

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$\large\rm{ 2 \sin^{2} \ y - \sin \ y = 0}$

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$\large\rm{ 2x^{2} - x = 0}$

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$\large\rm{ x(2x-1) = 0}$

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$\large\rm{ x = 0, x = \dfrac{1}{2}}$

but x = ½ doesn't satisfy the equation.

x ≠ ½.

so, $\large\rm{ \therefore x = 0}$