Question Number 29658 by gyugfeet last updated on 11/Feb/18
$${tan}\theta+{tan}\mathrm{2}\theta+\sqrt{\mathrm{3}}\:{tan}\theta.{tan}\mathrm{2}\theta=\sqrt{\mathrm{3}}\:\:\:\:\:\left(\mathrm{0}^{{o}\:} \leqslant\theta\leqslant\mathrm{360}\right) \\ $$
Answered by mrW2 last updated on 11/Feb/18
![tanθ+tan2θ=(√3)(1− tanθ.tan2θ) ((tan θ+tan 2θ)/(1−tan θ tan 2θ))=(√3) tan 3θ=(√3) 3θ=nπ+(π/3) θ=((nπ)/3)+(π/9) within [0,2π]: θ=(π/9),((4π)/9),((7π)/9),((10π)/9),((13π)/9),((16π)/9) or θ=20°,80°,140°,200°,260°,320°.](https://www.tinkutara.com/question/Q29664.png)
$${tan}\theta+{tan}\mathrm{2}\theta=\sqrt{\mathrm{3}}\left(\mathrm{1}−\:{tan}\theta.{tan}\mathrm{2}\theta\right) \\ $$$$\frac{\mathrm{tan}\:\theta+\mathrm{tan}\:\mathrm{2}\theta}{\mathrm{1}−\mathrm{tan}\:\theta\:\mathrm{tan}\:\mathrm{2}\theta}=\sqrt{\mathrm{3}} \\ $$$$\mathrm{tan}\:\mathrm{3}\theta=\sqrt{\mathrm{3}} \\ $$$$\mathrm{3}\theta={n}\pi+\frac{\pi}{\mathrm{3}} \\ $$$$\theta=\frac{{n}\pi}{\mathrm{3}}+\frac{\pi}{\mathrm{9}} \\ $$$${within}\:\left[\mathrm{0},\mathrm{2}\pi\right]: \\ $$$$\theta=\frac{\pi}{\mathrm{9}},\frac{\mathrm{4}\pi}{\mathrm{9}},\frac{\mathrm{7}\pi}{\mathrm{9}},\frac{\mathrm{10}\pi}{\mathrm{9}},\frac{\mathrm{13}\pi}{\mathrm{9}},\frac{\mathrm{16}\pi}{\mathrm{9}} \\ $$$${or} \\ $$$$\theta=\mathrm{20}°,\mathrm{80}°,\mathrm{140}°,\mathrm{200}°,\mathrm{260}°,\mathrm{320}°. \\ $$