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) )](Q54755.png)
$${solve}\:{for}\:{x} \\ $$$${sin}\left[\mathrm{2}{cos}^{−\mathrm{1}} \left\{{cot}\left(\mathrm{2}{tan}^{−\mathrm{1}} {x}\right)\right\}\right]=\mathrm{0} \\ $$$$ \\ $$$$\left({x}\:=\:\mathrm{1}\mp\sqrt{\mathrm{2}\:}\:,\:\mp\mathrm{1}\:,−\mathrm{1}\mp\sqrt{\mathrm{2}}\:\:\right) \\ $$
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](Q54758.png)
$$\theta={tan}^{−\mathrm{1}} {x}\rightarrow{tan}\theta={x} \\ $$$${cot}\left(\mathrm{2}\theta\right)=\frac{\mathrm{1}}{{tan}\mathrm{2}\theta}=\frac{\mathrm{1}−{tan}^{\mathrm{2}} \theta}{\mathrm{2}{tan}\theta}=\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{2}{x}} \\ $$$$ \\ $$$${now} \\ $$$${sin}\left[\mathrm{2}{cos}^{−\mathrm{1}} \left\{{cot}\left(\mathrm{2}{tan}^{−\mathrm{1}} {x}\right)\right\}\right]=\mathrm{0}\: \\ $$$${sin}\left[\mathrm{2}{cos}^{−\mathrm{1}} \left\{{cot}\left(\mathrm{2}{tan}^{−\mathrm{1}} {x}\right)\right\}\right]={sin}\mathrm{0}\:\:\:{or}\:{sin}\pi \\ $$$$\mathrm{2}{cos}^{−\mathrm{1}} \left\{{cot}\left(\mathrm{2}{tan}^{−\mathrm{1}} {x}\right)\right\}=\mathrm{0}\:\:\:{or}\:\:\pi \\ $$$${cot}\left(\mathrm{2}{tan}^{−\mathrm{1}} {x}\right)={cos}\left(\frac{\mathrm{0}}{\mathrm{2}}\right)\:\:\:{or}\:{cos}\left(\frac{\pi}{\mathrm{2}}\right) \\ $$$${cot}\left(\mathrm{2}{tan}^{−\mathrm{1}} {x}\right)=\mathrm{1}\:\:\:{or}\:\:\:\mathrm{0} \\ $$$$\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{2}{x}}=\mathrm{1}\:\:\:{or}\:\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{2}{x}}=\mathrm{0} \\ $$$${when} \\ $$$$\mathrm{1}−{x}^{\mathrm{2}} =\mathrm{2}{x} \\ $$$${x}^{\mathrm{2}} +\mathrm{2}{x}=\mathrm{1} \\ $$$$\left({x}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{2} \\ $$$$\left({x}+\mathrm{1}\right)=\pm\sqrt{\mathrm{2}}\: \\ $$$${x}=−\mathrm{1}\pm\sqrt{\mathrm{2}}\: \\ $$$${when}\:\:\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{2}{x}}=\mathrm{0}\rightarrow\mathrm{1}−{x}^{\mathrm{2}} =\mathrm{0}\rightarrow{x}^{\mathrm{2}} =\mathrm{1}\:\:{so}\:{x}=\pm\mathrm{1} \\ $$$$ \\ $$$$ \\ $$