solve for x sin[2cos^(−1) {cot(2tan^(−1) x)}]=0 (x = 1∓(√(2 )) , ∓1 ,−1∓(√2) )


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Question Number 54755 by Knight last updated on 10/Feb/19
$solve for x sin[2cos^(−1) {cot(2tan^(−1) x)}]=0 (x = 1∓(√(2 )) , ∓1 ,−1∓(√2) )$
$s o l v e f o r x$ $s i n [2 {c o s}^{- 1} {c o t (2 {t a n}^{- 1} x)}] = 0$ $(x = 1 \mp \sqrt{2}, \mp 1, - 1 \mp \sqrt{2})$
	Answered by tanmay.chaudhury50@gmail.com last updated on 10/Feb/19
	$θ=tan^(−1) x→tanθ=x cot(2θ)=(1/(tan2θ))=((1−tan^2 θ)/(2tanθ))=((1−x^2 )/(2x)) now sin[2cos^(−1) {cot(2tan^(−1) x)}]=0 sin[2cos^(−1) {cot(2tan^(−1) x)}]=sin0 or sinπ 2cos^(−1) {cot(2tan^(−1) x)}=0 or π cot(2tan^(−1) x)=cos((0/2)) or cos((π/2)) cot(2tan^(−1) x)=1 or 0 ((1−x^2 )/(2x))=1 or ((1−x^2 )/(2x))=0 when 1−x^2 =2x x^2 +2x=1 (x+1)^2 =2 (x+1)=±(√2) x=−1±(√2) when ((1−x^2 )/(2x))=0→1−x^2 =0→x^2 =1 so x=±1$
	$θ = {t a n}^{- 1} x \to t a n θ = x$ $c o t (2 θ) = \frac{1}{t a n 2 θ} = \frac{1 - {t a n}^{2} θ}{2 t a n θ} = \frac{1 - x^{2}}{2 x}$ $n o w$ $s i n [2 {c o s}^{- 1} {c o t (2 {t a n}^{- 1} x)}] = 0$ $s i n [2 {c o s}^{- 1} {c o t (2 {t a n}^{- 1} x)}] = s i n 0 o r s i n π$ $2 {c o s}^{- 1} {c o t (2 {t a n}^{- 1} x)} = 0 o r π$ $c o t (2 {t a n}^{- 1} x) = c o s (\frac{0}{2}) o r c o s (\frac{π}{2})$ $c o t (2 {t a n}^{- 1} x) = 1 o r 0$ $\frac{1 - x^{2}}{2 x} = 1 o r \frac{1 - x^{2}}{2 x} = 0$ $w h e n$ $1 - x^{2} = 2 x$ $x^{2} + 2 x = 1$ ${(x + 1)}^{2} = 2$ $(x + 1) = \pm \sqrt{2}$ $x = - 1 \pm \sqrt{2}$ $w h e n \frac{1 - x^{2}}{2 x} = 0 \to 1 - x^{2} = 0 \to x^{2} = 1 s o x = \pm 1$
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