Question

plzz find the value of x

Answer 1

hey mate
refer to attachment

Answer 2

Hi there!
Here's the answer:

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Given,
$sin^{-1}(1-x) - 2sin^{-1}x = \frac{\pi}{2}$

=> $sin^{-1}(1-x) = \frac{\pi}{2} + 2sin^{-1}x$

=> $(1-x) = sin(\frac{\pi}{2} + 2sin^{-1}x)$

=> $(1-x) = cos(2sin^{-1}x)$

=> $(1-x) = cos[cos^{-1}(1-2x^{2})]$

=> $(1-x) = 1 - 2x^{2}$

=> $2x^{2}-x = 0$

=> $x(2x-1) = 0$

=> x = 0 or x = $\frac{1}{2}$

• Check if x= $\frac{1}{2}$ satisfies the equation

LHS =

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

= $sin^{-1}(1-\frac{1}{2}) - 2sin^{-1}(\frac{1}{2})$

= $-sin^{-1}(\frac{1}{2})$

= $- \frac{\pi}{6}$

=/= RHS

• Check if x= 0 satisfies the equation

LHS =

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

= $sin^{-1}(1-0) - 2sin^{-1}(0)$

= $\frac{\pi}{2} - 0$

= $\frac{\pi}{2}$

= RHS

•°• x = 0 is the required solution



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