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Question Number 171872 by Shrinava last updated on 21/Jun/22

Solve for real numbers: (1/(1 + tan^4 x)) + (1/(10)) = (2/(1 + 3 tan^2 x))

$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{tan}^{\mathrm{4}} \:\mathrm{x}}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{10}}\:\:=\:\:\frac{\mathrm{2}}{\mathrm{1}\:+\:\mathrm{3}\:\mathrm{tan}^{\mathrm{2}} \:\mathrm{x}} \\ $$

Answered by aleks041103 last updated on 21/Jun/22

t=tan^2 x (1/(1+t^2 ))+(1/(10))=(2/(1+3t)) (1+3t)+(((1+3t)(1+t^2 ))/(10))=2(1+t^2 ) 10+30t+1+3t+t^2 +3t^3 =20+20t^2 3t^3 −19t^2 +33t−9=0, t>0 3t^3 −t^2 −18t^2 +33t−9=0 t^2 (3t−1)−18t^2 +6t+27t−9=0 t^2 (3t−1)−6t(3t−1)+9(3t−1)=0 (3t−1)(t^2 −6t+9)=0 (3t−1)(t−3)^2 =0 t=(1/3),3 tgx=±((√3)/3),±(√3) x=±(π/6)+kπ;±(π/3)+kπ

$${t}={tan}^{\mathrm{2}} {x} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{10}}=\frac{\mathrm{2}}{\mathrm{1}+\mathrm{3}{t}} \\ $$$$\left(\mathrm{1}+\mathrm{3}{t}\right)+\frac{\left(\mathrm{1}+\mathrm{3}{t}\right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{\mathrm{10}}=\mathrm{2}\left(\mathrm{1}+{t}^{\mathrm{2}} \right) \\ $$$$\mathrm{10}+\mathrm{30}{t}+\mathrm{1}+\mathrm{3}{t}+{t}^{\mathrm{2}} +\mathrm{3}{t}^{\mathrm{3}} =\mathrm{20}+\mathrm{20}{t}^{\mathrm{2}} \\ $$$$\mathrm{3}{t}^{\mathrm{3}} −\mathrm{19}{t}^{\mathrm{2}} +\mathrm{33}{t}−\mathrm{9}=\mathrm{0},\:{t}>\mathrm{0} \\ $$$$\mathrm{3}{t}^{\mathrm{3}} −{t}^{\mathrm{2}} −\mathrm{18}{t}^{\mathrm{2}} +\mathrm{33}{t}−\mathrm{9}=\mathrm{0} \\ $$$${t}^{\mathrm{2}} \left(\mathrm{3}{t}−\mathrm{1}\right)−\mathrm{18}{t}^{\mathrm{2}} +\mathrm{6}{t}+\mathrm{27}{t}−\mathrm{9}=\mathrm{0} \\ $$$${t}^{\mathrm{2}} \left(\mathrm{3}{t}−\mathrm{1}\right)−\mathrm{6}{t}\left(\mathrm{3}{t}−\mathrm{1}\right)+\mathrm{9}\left(\mathrm{3}{t}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\left(\mathrm{3}{t}−\mathrm{1}\right)\left({t}^{\mathrm{2}} −\mathrm{6}{t}+\mathrm{9}\right)=\mathrm{0} \\ $$$$\left(\mathrm{3}{t}−\mathrm{1}\right)\left({t}−\mathrm{3}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${t}=\frac{\mathrm{1}}{\mathrm{3}},\mathrm{3} \\ $$$${tgx}=\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{3}},\pm\sqrt{\mathrm{3}} \\ $$$${x}=\pm\frac{\pi}{\mathrm{6}}+{k}\pi;\pm\frac{\pi}{\mathrm{3}}+{k}\pi \\ $$

Commented by Shrinava last updated on 22/Jun/22

Cool dear professor thank you

$$\mathrm{Cool}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{thank}\:\mathrm{you} \\ $$

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