0= 2 ^ -1 -x 0= ( ^ -1 (-2 1-x^2 ) ) ..... [2 ^ -1 x= ^ -1 (2 1-x^2 ) ] 0= ( ^ -1 1- (4x^2-4x^4 ) ) ..... [ ^ -1 x= ^ -1 1-x^2 ] 0= 1- (4x^2-4x^4 ) 0=1- (4x^2-4x^4 ) 4x^4-4x^2+1=0 4x^4-2x^2-2x^2+1=0 2x^2 (2x^2-1 )-1 (2x^2-1 )=0 (2x^2-1 )^2=0 2x^2-1=0 x= 1 2

Solve: 2 ^ -1 -x =0

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Question

Solve: $\cos\Big\{2\sin^{-1}\{-\text{x}\}\Big\}=0$

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Answer

$0=\cos\Big\{2\sin^{-1}\{-\text{x}\}\Big\}$
$0=\cos\Big(\sin^{-1}\Big(-2\times\sqrt{1-\text{x}^2}\Big)\Big)\\.....\Big[2\sin^{-1}\text{x}=\sin^{-1}\Big(2\times\sqrt{1-\text{x}^2}\Big)\Big ]$
$$$0=\cos\Big(\cos^{-1}\sqrt{1-\big(4\text{x}^2-4\text{x}^4}\big)\Big)\\.....\Big[\sin^{-1}\text{x}=\cos^{-1}\sqrt{1-\text{x}^2}\Big]$
$$$0=\sqrt{1-\big(4\text{x}^2-4\text{x}^4\big)}$
$0=1-\big(4\text{x}^2-4\text{x}^4\big)$
$4\text{x}^4-4\text{x}^2+1=0$
$4\text{x}^4-2\text{x}^2-2\text{x}^2+1=0$
$2\text{x}^2\big(2\text{x}^2-1\big)-1\big(2\text{x}^2-1\big)=0$
$\big(2\text{x}^2-1\big)^2=0$
$2\text{x}^2-1=0$
$\text{x}=\pm\frac{1}{\sqrt2}$

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