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Question Number 85372 by john santu last updated on 21/Mar/20

sin^2 x−4cos^2 x = 3 sin x cos x  find tan x.

$$\mathrm{sin}\:^{\mathrm{2}} {x}−\mathrm{4cos}\:^{\mathrm{2}} {x}\:=\:\mathrm{3}\:\mathrm{sin}\:{x}\:\mathrm{cos}\:{x} \\ $$$${find}\:\mathrm{tan}\:{x}.\: \\ $$

Answered by john santu last updated on 21/Mar/20

tan^2 x − 4 = 3 tan x  tan^2 x − 3tan x − 4 = 0  tan x = ((3 ± 5)/2) ⇒  { ((tan x = 4)),((tan x = −1)) :}

$$\mathrm{tan}\:^{\mathrm{2}} {x}\:−\:\mathrm{4}\:=\:\mathrm{3}\:\mathrm{tan}\:{x} \\ $$$$\mathrm{tan}\:^{\mathrm{2}} {x}\:−\:\mathrm{3tan}\:{x}\:−\:\mathrm{4}\:=\:\mathrm{0} \\ $$$$\mathrm{tan}\:{x}\:=\:\frac{\mathrm{3}\:\pm\:\mathrm{5}}{\mathrm{2}}\:\Rightarrow\:\begin{cases}{\mathrm{tan}\:{x}\:=\:\mathrm{4}}\\{\mathrm{tan}\:{x}\:=\:−\mathrm{1}}\end{cases} \\ $$

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