solve : tan x−tan (2x) = 2(√3)


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

$solve : \tan x - \tan (2 x) = 2 \sqrt{3}$
	Answered by bobhans last updated on 04/Jun/20
	$⇒ tan (x)−((2tan (x))/(1−tan^2 (x))) = 2(√3) tan (x)−tan^3 (x)−2tan (x)= 2(√3) −2(√3) tan^2 (x) let tan (x) = v ⇒v^3 −2(√3) v^2 + v+ 2(√3) = 0 (v−(√3)) (v^2 −(√3)v −2) = 0 { ((v = (√3) )),((v = (((√3) + (√(11)))/2))),((v = (((√3) −(√(11)))/2))) :} { ((tan (x) = (√3) ⇒x = (π/3)+ nπ)),((tan (x) = (((√3)+(√(11)))/2) ⇒ x = arctan ((((√3)+(√(11)))/2))+nπ)),((tan (x) = (((√3)−(√(11)))/2) ⇒x = arctan ((((√3)−(√(11)))/2)) +nπ)) :}$
	$\Rightarrow \tan (x) - \frac{2 \tan (x)}{1 - \tan^{2} (x)} = 2 \sqrt{3}$ $\tan (x) - \tan^{3} (x) - 2 \tan (x) = 2 \sqrt{3} - 2 \sqrt{3} \tan^{2} (x)$ $l e t \tan (x) = v \Rightarrow v^{3} - 2 \sqrt{3} v^{2} + v + 2 \sqrt{3} = 0$ $(v - \sqrt{3}) (v^{2} - \sqrt{3} v - 2) = 0$ ${\begin{cases} v = \sqrt{3} \\ v = \frac{\sqrt{3} + \sqrt{11}}{2} \\ v = \frac{\sqrt{3} - \sqrt{11}}{2} \end{cases}$ ${\begin{cases} \tan (x) = \sqrt{3} \Rightarrow x = \frac{π}{3} + n π \\ \tan (x) = \frac{\sqrt{3} + \sqrt{11}}{2} \Rightarrow x = \arctan (\frac{\sqrt{3} + \sqrt{11}}{2}) + n π \\ \tan (x) = \frac{\sqrt{3} - \sqrt{11}}{2} \Rightarrow x = \arctan (\frac{\sqrt{3} - \sqrt{11}}{2}) + n π \end{cases}$
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