Question Number 24294 by ajfour last updated on 15/Nov/17
![Find value(s) of x if sin [2cos^(−1) {cot (2tan^(−1) x)}]=0 .](https://www.tinkutara.com/question/Q24294.png)
$${Find}\:{value}\left({s}\right)\:{of}\:\boldsymbol{{x}}\:{if} \\ $$$$\:\:\mathrm{sin}\:\left[\mathrm{2cos}^{−\mathrm{1}} \left\{\mathrm{cot}\:\left(\mathrm{2tan}^{−\mathrm{1}} {x}\right)\right\}\right]=\mathrm{0}\:. \\ $$
Answered by mrW1 last updated on 15/Nov/17
![sin [2cos^(−1) {cot (2tan^(−1) x)}]=0 ⇒2cos^(−1) {cot (2tan^(−1) x)}=kπ ⇒cos^(−1) {cot (2tan^(−1) x)}=((kπ)/2), k∈[0,2] ⇒cot (2tan^(−1) x)=cos (((kπ)/2)), k∈[0,2] ⇒cot (2tan^(−1) x)=±1, 0 ⇒2tan^(−1) x=kπ±(π/4), kπ+(π/2) ⇒tan^(−1) x=−((3π)/8),−(π/4), −(π/8), (π/8),(π/4), ((3π)/8) ⇒x=±tan ((π/8))=±((√2)−1) ⇒x=±tan ((π/4))=±1 ⇒x=±tan (((3π)/8))=±((√2)+1) i.e. there are 6 values for x.](https://www.tinkutara.com/question/Q24310.png)
$$\:\mathrm{sin}\:\left[\mathrm{2cos}^{−\mathrm{1}} \left\{\mathrm{cot}\:\left(\mathrm{2tan}^{−\mathrm{1}} {x}\right)\right\}\right]=\mathrm{0}\: \\ $$$$\:\Rightarrow\mathrm{2cos}^{−\mathrm{1}} \left\{\mathrm{cot}\:\left(\mathrm{2tan}^{−\mathrm{1}} {x}\right)\right\}={k}\pi\: \\ $$$$\:\Rightarrow\mathrm{cos}^{−\mathrm{1}} \left\{\mathrm{cot}\:\left(\mathrm{2tan}^{−\mathrm{1}} {x}\right)\right\}=\frac{{k}\pi}{\mathrm{2}},\:{k}\in\left[\mathrm{0},\mathrm{2}\right] \\ $$$$\:\Rightarrow\mathrm{cot}\:\left(\mathrm{2tan}^{−\mathrm{1}} {x}\right)=\mathrm{cos}\:\left(\frac{{k}\pi}{\mathrm{2}}\right),\:{k}\in\left[\mathrm{0},\mathrm{2}\right] \\ $$$$\:\Rightarrow\mathrm{cot}\:\left(\mathrm{2tan}^{−\mathrm{1}} {x}\right)=\pm\mathrm{1},\:\mathrm{0} \\ $$$$\:\Rightarrow\mathrm{2tan}^{−\mathrm{1}} {x}={k}\pi\pm\frac{\pi}{\mathrm{4}},\:{k}\pi+\frac{\pi}{\mathrm{2}} \\ $$$$\:\Rightarrow\mathrm{tan}^{−\mathrm{1}} {x}=−\frac{\mathrm{3}\pi}{\mathrm{8}},−\frac{\pi}{\mathrm{4}},\:−\frac{\pi}{\mathrm{8}},\:\frac{\pi}{\mathrm{8}},\frac{\pi}{\mathrm{4}},\:\frac{\mathrm{3}\pi}{\mathrm{8}} \\ $$$$\Rightarrow{x}=\pm\mathrm{tan}\:\left(\frac{\pi}{\mathrm{8}}\right)=\pm\left(\sqrt{\mathrm{2}}−\mathrm{1}\right) \\ $$$$\Rightarrow{x}=\pm\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}\right)=\pm\mathrm{1} \\ $$$$\Rightarrow{x}=\pm\mathrm{tan}\:\left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)=\pm\left(\sqrt{\mathrm{2}}+\mathrm{1}\right) \\ $$$${i}.{e}.\:{there}\:{are}\:\mathrm{6}\:{values}\:{for}\:{x}. \\ $$
Commented by ajfour last updated on 15/Nov/17
$${yes}\:{sir},\:{thank}\:{you}. \\ $$