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Question Number 207724 by hardmath last updated on 24/May/24

2 tg^3 x βˆ’ 2 tg^2 x + 6 tg x = 3 , [0 ; 2𝛑] Sum of roots = ?

$$\mathrm{2}\:\mathrm{tg}^{\mathrm{3}} \:\boldsymbol{\mathrm{x}}\:βˆ’\:\mathrm{2}\:\mathrm{tg}^{\mathrm{2}} \:\boldsymbol{\mathrm{x}}\:+\:\mathrm{6}\:\mathrm{tg}\:\boldsymbol{\mathrm{x}}\:=\:\mathrm{3}\:\:\:,\:\:\:\left[\mathrm{0}\:;\:\mathrm{2}\boldsymbol{\pi}\right] \\ $$$$\mathrm{Sum}\:\mathrm{of}\:\mathrm{roots}\:=\:? \\ $$

Answered by Frix last updated on 24/May/24

tan^3 x βˆ’tan^2 x +3tan x βˆ’(3/2)=0 t=tan x t^3 βˆ’t^2 +3tβˆ’(3/3)=0 t=z+(1/3) z^3 +((8z)/3)βˆ’((31)/(54))=0 z=(((31)/(108))+((√(113))/(12)))^(1/3) βˆ’(βˆ’((31)/(108))+((√(113))/(12)))^(1/3) t=(1/3)+(((31)/(108))+((√(113))/(12)))^(1/3) βˆ’(βˆ’((31)/(108))+((√(113))/(12)))^(1/3) x_1 =tan^(βˆ’1) ((1/3)+(((31)/(108))+((√(113))/(12)))^(1/3) βˆ’(βˆ’((31)/(108))+((√(113))/(12)))^(1/3) ) x_2 =x_1 +Ο€ Sum of roots is Ο€+2tan^(βˆ’1) ((1/3)+(((31)/(108))+((√(113))/(12)))^(1/3) βˆ’(βˆ’((31)/(108))+((√(113))/(12)))^(1/3) ) Not very nice but true anyway...

$$\mathrm{tan}^{\mathrm{3}} \:{x}\:βˆ’\mathrm{tan}^{\mathrm{2}} \:{x}\:+\mathrm{3tan}\:{x}\:βˆ’\frac{\mathrm{3}}{\mathrm{2}}=\mathrm{0} \\ $$$${t}=\mathrm{tan}\:{x} \\ $$$${t}^{\mathrm{3}} βˆ’{t}^{\mathrm{2}} +\mathrm{3}{t}βˆ’\frac{\mathrm{3}}{\mathrm{3}}=\mathrm{0} \\ $$$${t}={z}+\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${z}^{\mathrm{3}} +\frac{\mathrm{8}{z}}{\mathrm{3}}βˆ’\frac{\mathrm{31}}{\mathrm{54}}=\mathrm{0} \\ $$$${z}=\left(\frac{\mathrm{31}}{\mathrm{108}}+\frac{\sqrt{\mathrm{113}}}{\mathrm{12}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} βˆ’\left(βˆ’\frac{\mathrm{31}}{\mathrm{108}}+\frac{\sqrt{\mathrm{113}}}{\mathrm{12}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$${t}=\frac{\mathrm{1}}{\mathrm{3}}+\left(\frac{\mathrm{31}}{\mathrm{108}}+\frac{\sqrt{\mathrm{113}}}{\mathrm{12}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} βˆ’\left(βˆ’\frac{\mathrm{31}}{\mathrm{108}}+\frac{\sqrt{\mathrm{113}}}{\mathrm{12}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$${x}_{\mathrm{1}} =\mathrm{tan}^{βˆ’\mathrm{1}} \:\left(\frac{\mathrm{1}}{\mathrm{3}}+\left(\frac{\mathrm{31}}{\mathrm{108}}+\frac{\sqrt{\mathrm{113}}}{\mathrm{12}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} βˆ’\left(βˆ’\frac{\mathrm{31}}{\mathrm{108}}+\frac{\sqrt{\mathrm{113}}}{\mathrm{12}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \right) \\ $$$${x}_{\mathrm{2}} ={x}_{\mathrm{1}} +\pi \\ $$$$\mathrm{Sum}\:\mathrm{of}\:\mathrm{roots}\:\mathrm{is} \\ $$$$\pi+\mathrm{2tan}^{βˆ’\mathrm{1}} \:\left(\frac{\mathrm{1}}{\mathrm{3}}+\left(\frac{\mathrm{31}}{\mathrm{108}}+\frac{\sqrt{\mathrm{113}}}{\mathrm{12}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} βˆ’\left(βˆ’\frac{\mathrm{31}}{\mathrm{108}}+\frac{\sqrt{\mathrm{113}}}{\mathrm{12}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \right) \\ $$$$\mathrm{Not}\:\mathrm{very}\:\mathrm{nice}\:\mathrm{but}\:\mathrm{true}\:\mathrm{anyway}... \\ $$

Commented by hardmath last updated on 25/May/24

thank you dear professor

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

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