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Question-194654




Question Number 194654 by sonukgindia last updated on 12/Jul/23
Answered by MM42 last updated on 12/Jul/23
((3+2(√2))^(tanx) )^(tanx+2) =u  ⇒u+(1/u)=6⇒u^2 −6u+1=0  ⇒u=3±2(√2)  1)⇒((3+2(√2))^(tanx) )^(tanx+2) =3+2(√2)  ⇒tan^2 x+2tanx=1⇒tanx=−1±(√2)  ⇒x=tan^(−1) (−1±(√(2 )))=22.5 or −67.5  2)((3+2(√2))^(tanx) )^(tanx+2) =3−2(√2)  tan^2 x+2tanx=−1⇒tanx=±1  ⇒x=±45
$$\left(\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)^{{tanx}} \right)^{{tanx}+\mathrm{2}} ={u} \\ $$$$\Rightarrow{u}+\frac{\mathrm{1}}{{u}}=\mathrm{6}\Rightarrow{u}^{\mathrm{2}} −\mathrm{6}{u}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{u}=\mathrm{3}\pm\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\left.\mathrm{1}\right)\Rightarrow\left(\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)^{{tanx}} \right)^{{tanx}+\mathrm{2}} =\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\Rightarrow{tan}^{\mathrm{2}} {x}+\mathrm{2}{tanx}=\mathrm{1}\Rightarrow{tanx}=−\mathrm{1}\pm\sqrt{\mathrm{2}} \\ $$$$\Rightarrow{x}={tan}^{−\mathrm{1}} \left(−\mathrm{1}\pm\sqrt{\mathrm{2}\:}\right)=\mathrm{22}.\mathrm{5}\:{or}\:−\mathrm{67}.\mathrm{5} \\ $$$$\left.\mathrm{2}\right)\left(\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)^{{tanx}} \right)^{{tanx}+\mathrm{2}} =\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}} \\ $$$${tan}^{\mathrm{2}} {x}+\mathrm{2}{tanx}=−\mathrm{1}\Rightarrow{tanx}=\pm\mathrm{1} \\ $$$$\Rightarrow{x}=\pm\mathrm{45} \\ $$$$ \\ $$

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