Question

$\bf \: {sin}^{ - 1} (1 - x) - 2 {sin}^{ - 1} x = \frac{\pi}{2} \: \:$

Then the value of x is equal ​

Answer 1

Hence, x=0 is the only solution of the equation.

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Answer 2

Answer:

$sin^{-1}(1-x)-2sin^{-1}x=\dfrac{\pi}{2}$

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

$\text{Taking sine both sides we get :}\\\\\implies 1-x=sin(\dfrac{\pi}{2}+2sin^{-1}x})\\\\\implies 1-x=sin(\dfrac{\pi}{2}-(-2sin^{-1}{x}))\\\\\implies 1-x=cos(2sin^{-1}x)\\\\\mathsf{Use\:cos(2sin^{-1}x)=1-2x^2}\\\implies 1-x=1-2x^2\\\\\implies 2x^2-x=0\\\\\implies x(2x-1)=0\\\\\implies x=0\:or\:x=\dfrac{1}{2}$

Answer is 0 .

Step-by-step explanation:

Note the step :

cos ( 2 sin⁻¹ x ) = 1 - 2 x² .

cos ( 2 sin⁻¹ x )  = 1 - 2 [ sin ⁻¹x ( sin x )  ]²

= 1 - 2 x²

We neglect 1/2 because LHS is not equal to RHS when x = 1/2 and hence the correct value will be 0 .