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Question Number 112145 by weltr last updated on 06/Sep/20

4^(tan^2 x)  + 2^(1/(cos^2 x))  − 80 = 0

$$\mathrm{4}^{\mathrm{tan}\:^{\mathrm{2}} {x}} \:+\:\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{cos}\:^{\mathrm{2}} {x}}} \:−\:\mathrm{80}\:=\:\mathrm{0} \\ $$

Answered by bemath last updated on 06/Sep/20

4^(tan^2 x)  + 2^(sec^2 x)  − 80 = 0  4^(tan^2 x)  + 2^(tan^2 x+1)  −80 = 0  set 2^(tan^2 x)  = q  ⇒ q^2 +2q−80 = 0  ⇒(q+10)(q−8) = 0  ⇒q = 8 ⇒ 2^(tan^2 x)  = 2^3  ⇒ tan x = ± (√3)

$$\mathrm{4}^{\mathrm{tan}\:^{\mathrm{2}} {x}} \:+\:\mathrm{2}^{\mathrm{sec}\:^{\mathrm{2}} {x}} \:−\:\mathrm{80}\:=\:\mathrm{0} \\ $$$$\mathrm{4}^{\mathrm{tan}\:^{\mathrm{2}} {x}} \:+\:\mathrm{2}^{\mathrm{tan}\:^{\mathrm{2}} {x}+\mathrm{1}} \:−\mathrm{80}\:=\:\mathrm{0} \\ $$$${set}\:\mathrm{2}^{\mathrm{tan}\:^{\mathrm{2}} {x}} \:=\:{q} \\ $$$$\Rightarrow\:{q}^{\mathrm{2}} +\mathrm{2}{q}−\mathrm{80}\:=\:\mathrm{0} \\ $$$$\Rightarrow\left({q}+\mathrm{10}\right)\left({q}−\mathrm{8}\right)\:=\:\mathrm{0} \\ $$$$\Rightarrow{q}\:=\:\mathrm{8}\:\Rightarrow\:\mathrm{2}^{\mathrm{tan}\:^{\mathrm{2}} {x}} \:=\:\mathrm{2}^{\mathrm{3}} \:\Rightarrow\:\mathrm{tan}\:{x}\:=\:\pm\:\sqrt{\mathrm{3}} \\ $$

Commented by weltr last updated on 06/Sep/20

is this the final answer?

$${is}\:{this}\:{the}\:{final}\:{answer}? \\ $$

Commented by bobhans last updated on 06/Sep/20

tan x = (√3) → x = 60°+k.180°  tan x = −(√3) →x=−60°+k.180° ; k∈Z

$$\mathrm{tan}\:\mathrm{x}\:=\:\sqrt{\mathrm{3}}\:\rightarrow\:\mathrm{x}\:=\:\mathrm{60}°+\mathrm{k}.\mathrm{180}° \\ $$$$\mathrm{tan}\:\mathrm{x}\:=\:−\sqrt{\mathrm{3}}\:\rightarrow\mathrm{x}=−\mathrm{60}°+\mathrm{k}.\mathrm{180}°\:;\:\mathrm{k}\in\mathbb{Z} \\ $$

Commented by weltr last updated on 07/Sep/20

thanks

$${thanks} \\ $$

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