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Question Number 194654 by sonukgindia last updated on 12/Jul/23

Answered by MM42 last updated on 12/Jul/23

$${\left({(3+2\sqrt{2})}^{tanx}\right)}^{tanx+2}=u$$$$\Rightarrow u+\frac{1}{u}=6\Rightarrow u^2-6u+1=0$$$$\Rightarrow u=3\pm 2\sqrt{2}$$$$1)\Rightarrow {\left({(3+2\sqrt{2})}^{tanx}\right)}^{tanx+2}=3+2\sqrt{2}$$$$\Rightarrow {tan}^{2}x+2tanx=1\Rightarrow tanx=-1\pm \sqrt{2}$$$$\Rightarrow x={tan}^{-1}(-1\pm \sqrt{2})=22.5or-67.5$$$$2){\left({(3+2\sqrt{2})}^{tanx}\right)}^{tanx+2}=3-2\sqrt{2}$$$${tan}^{2}x+2tanx=-1\Rightarrow tanx=\pm 1$$$$\Rightarrow x=\pm 45$$$$$$

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