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Question Number 120467 by help last updated on 31/Oct/20

Answered by mathmax by abdo last updated on 31/Oct/20

$$\tan (\theta +\frac{\pi }{3})=\frac{\tan \theta +\tan \frac{\pi }{3}}{1-\tan \theta \tan \left(\frac{\pi }{3}\right)}=\frac{\tan \theta +\sqrt{3}}{1-\sqrt{3}\tan \theta }$$$$\tan (\theta +\frac{2\pi }{3})=\tan (\theta +\pi -\frac{\pi }{3})=\frac{\tan \theta -\sqrt{3}}{1+\sqrt{3}\tan \theta }$$$$\mathrm{e}\Rightarrow \tan \theta +\frac{\tan \theta +\sqrt{3}}{1-\sqrt{3}\tan \theta }+\frac{\tan \theta -\sqrt{3}}{1+\sqrt{3}\mathrm{tant}}=3\mathrm{let}\tan \theta =\mathrm{x}\mathrm{so}\mathrm{e}\Rightarrow$$$$\mathrm{x}+\frac{\mathrm{x}+\sqrt{3}}{1-\sqrt{3}\mathrm{x}}+\frac{\mathrm{x}-\sqrt{3}}{1+\sqrt{3}\mathrm{x}}=3\Rightarrow \mathrm{x}+\frac{(\mathrm{x}+\sqrt{3})(1+\sqrt{3}\mathrm{x})+(\mathrm{x}-\sqrt{3})(1-\sqrt{3}\mathrm{x})}{1-{3\mathrm{x}}^2}=3$$$$\Rightarrow \mathrm{x}+\frac{\mathrm{x}+\sqrt{3}{\mathrm{x}}^2+\sqrt{3}+3\mathrm{x}+\mathrm{x}-\sqrt{3}{\mathrm{x}}^2-\sqrt{3}+3\mathrm{x}}{1-{3\mathrm{x}}^2}=3\Rightarrow$$$$\mathrm{x}+\frac{8\mathrm{x}}{1-{3\mathrm{x}}^2}=3\Rightarrow \mathrm{x}-{3\mathrm{x}}^3+8\mathrm{x}=3-{9\mathrm{x}}^2\Rightarrow$$$$9\mathrm{x}-{3\mathrm{x}}^3-3+{9\mathrm{x}}^2=0\Rightarrow 3\mathrm{x}-{\mathrm{x}}^3-1+{3\mathrm{x}}^2=0\Rightarrow$$$$-{\mathrm{x}}^3+{3\mathrm{x}}^2+3\mathrm{x}-1=0\Rightarrow {\mathrm{x}}^3-{3\mathrm{x}}^2-3\mathrm{x}+1=0-1\mathrm{is}\mathrm{roots}$$$$\Rightarrow {\mathrm{x}}^3-{3\mathrm{x}}^2-3\mathrm{x}+1=(\mathrm{x}+1)({\mathrm{x}}^2+\mathrm{ax}+\mathrm{b})$$$$\mathrm{b}=1\Rightarrow (\mathrm{x}+1)({\mathrm{x}}^2+\mathrm{ax}+1)={\mathrm{x}}^3-{3\mathrm{x}}^2-3\mathrm{x}+1\Rightarrow$$$${\mathrm{x}}^3+{\mathrm{ax}}^2+\mathrm{x}+{\mathrm{x}}^2+\mathrm{ax}+1={\mathrm{x}}^3-{3\mathrm{x}}^2-3\mathrm{x}+1\Rightarrow$$$$(\mathrm{a}+1){\mathrm{x}}^2+(\mathrm{a}+1)\mathrm{x}+1=-{3\mathrm{x}}^2-3\mathrm{x}+1\Rightarrow \mathrm{a}+1=-3\Rightarrow \mathrm{a}=-4\Rightarrow$$$${\mathrm{x}}^3-{3\mathrm{x}}^2-3\mathrm{x}+1=0\Rightarrow (\mathrm{x}+1)({\mathrm{x}}^2-4\mathrm{x}+1)=0\Rightarrow \mathrm{x}=-1\mathrm{or}$$$${\mathrm{x}}^2-4\mathrm{x}+1=0$$$$\mathrm{tanx}=-1\Rightarrow \mathrm{tanx}=\tan (-\frac{\pi }{4})\Rightarrow \mathrm{x}=-\frac{\pi }{4}+\mathrm{k}\pi (\mathrm{k}\in \mathrm{Z})$$$${\mathrm{x}}^2-4\mathrm{x}+1=0\to {\mathrm{\Delta }}^{{}^{\prime }}=4-1=3\Rightarrow {\mathrm{x}}_1=2+\sqrt{3}\mathrm{and}{\mathrm{x}}_2=2-\sqrt{3}$$$$\mathrm{tanx}=2+\sqrt{3}\Rightarrow \mathrm{x}=\arctan (2+\sqrt{3})+\mathrm{K}\pi$$$$\mathrm{tanx}=2-\sqrt{3}\Rightarrow \mathrm{x}=\arctan (2-\sqrt{3})+\mathrm{k}\pi$$

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