Solve: 2 ^ -1 -x =0 - Vidyadip

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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