Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

Question Number 96766 by bemath last updated on 04/Jun/20

solve : tan x−tan (2x) = 2(√3)

$$\mathrm{solve}\::\:\mathrm{tan}\:{x}−\mathrm{tan}\:\left(\mathrm{2}{x}\right)\:=\:\mathrm{2}\sqrt{\mathrm{3}}\: \\ $$

Answered by bobhans last updated on 04/Jun/20

⇒ tan (x)−((2tan (x))/(1−tan^2 (x))) = 2(√3)  tan (x)−tan^3 (x)−2tan (x)= 2(√3) −2(√3) tan^2 (x)  let tan (x) = v ⇒v^3 −2(√3) v^2 + v+ 2(√3) = 0  (v−(√3)) (v^2 −(√3)v −2) = 0   { ((v = (√3) )),((v = (((√3) + (√(11)))/2))),((v = (((√3) −(√(11)))/2))) :}   { ((tan (x) = (√3) ⇒x = (π/3)+ nπ)),((tan (x) = (((√3)+(√(11)))/2) ⇒ x = arctan ((((√3)+(√(11)))/2))+nπ)),((tan (x) = (((√3)−(√(11)))/2) ⇒x = arctan ((((√3)−(√(11)))/2)) +nπ)) :}

$$\Rightarrow\:\mathrm{tan}\:\left({x}\right)−\frac{\mathrm{2tan}\:\left({x}\right)}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \left({x}\right)}\:=\:\mathrm{2}\sqrt{\mathrm{3}} \\ $$$$\mathrm{tan}\:\left({x}\right)−\mathrm{tan}\:^{\mathrm{3}} \left({x}\right)−\mathrm{2tan}\:\left({x}\right)=\:\mathrm{2}\sqrt{\mathrm{3}}\:−\mathrm{2}\sqrt{\mathrm{3}}\:\mathrm{tan}\:^{\mathrm{2}} \left({x}\right) \\ $$$${let}\:\mathrm{tan}\:\left({x}\right)\:=\:{v}\:\Rightarrow{v}^{\mathrm{3}} −\mathrm{2}\sqrt{\mathrm{3}}\:{v}^{\mathrm{2}} +\:{v}+\:\mathrm{2}\sqrt{\mathrm{3}}\:=\:\mathrm{0} \\ $$$$\left({v}−\sqrt{\mathrm{3}}\right)\:\left({v}^{\mathrm{2}} −\sqrt{\mathrm{3}}{v}\:−\mathrm{2}\right)\:=\:\mathrm{0} \\ $$$$\begin{cases}{{v}\:=\:\sqrt{\mathrm{3}}\:}\\{{v}\:=\:\frac{\sqrt{\mathrm{3}}\:+\:\sqrt{\mathrm{11}}}{\mathrm{2}}}\\{{v}\:=\:\frac{\sqrt{\mathrm{3}}\:−\sqrt{\mathrm{11}}}{\mathrm{2}}}\end{cases} \\ $$$$\begin{cases}{\mathrm{tan}\:\left({x}\right)\:=\:\sqrt{\mathrm{3}}\:\Rightarrow{x}\:=\:\frac{\pi}{\mathrm{3}}+\:\mathrm{n}\pi}\\{\mathrm{tan}\:\left({x}\right)\:=\:\frac{\sqrt{\mathrm{3}}+\sqrt{\mathrm{11}}}{\mathrm{2}}\:\Rightarrow\:{x}\:=\:\mathrm{arctan}\:\left(\frac{\sqrt{\mathrm{3}}+\sqrt{\mathrm{11}}}{\mathrm{2}}\right)+{n}\pi}\\{\mathrm{tan}\:\left({x}\right)\:=\:\frac{\sqrt{\mathrm{3}}−\sqrt{\mathrm{11}}}{\mathrm{2}}\:\Rightarrow{x}\:=\:\mathrm{arctan}\:\left(\frac{\sqrt{\mathrm{3}}−\sqrt{\mathrm{11}}}{\mathrm{2}}\right)\:+{n}\pi}\end{cases} \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com