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tan-3x-tan-5x-2-Find-x-

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Question Number 209735 by hardmath last updated on 19/Jul/24
tan(3x) + tan(5x) = 2 Find: x = ?
$$\mathrm{tan}\left(\mathrm{3x}\right)\:\:+\:\:\mathrm{tan}\left(\mathrm{5x}\right)\:\:=\:\:\mathrm{2} \\ $$$$\mathrm{Find}:\:\:\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$
Answered by a.lgnaoui last updated on 20/Jul/24
voir reponse en bas (a l exeption d ′ereur de calcul)
$$\mathrm{voir}\:\mathrm{reponse}\:\mathrm{en}\:\mathrm{bas} \\ $$$$\:\left(\mathrm{a}\:\mathrm{l}\:\mathrm{exeption}\:\:\mathrm{d}\:'\mathrm{ereur}\:\mathrm{de}\:\mathrm{calcul}\right) \\ $$
Commented by a.lgnaoui last updated on 20/Jul/24
Commented by a.lgnaoui last updated on 20/Jul/24
Answered by a.lgnaoui last updated on 21/Jul/24
{ ((tan a+tan b=((tan (a+b))/(1−tan atan b)))),((avec t=tan x tan 2x=((2t)/(2−t^2 )))) :} 2=((tan 8x)/(1−tan 3x.tan 5x)) •tan 3x=((tan 2x+tan x)/(1−tan 2x.tan x))=((t(3−t^2 ))/(1−3t^2 )) tan 3x=(((3−t)t)/((1−3t^2 )) (1) •tan 5x=((tan 2x+tan 3x)/(1−tan 2x.tan 3x)) =((2t(1−3t^2 )+t(1−t^2 )(3−t^2 ))/((1−t^2 )(1−3t^2 )−2t^2 (3−t^2 ))) tan 5x=((t(t^4 −10t^2 +5))/(5t^4 −10t^2 +1)) (2) 2=tan 3x+tan 5x ⇒ 2=(((3−t)t)/(1−3t^2 ))+((t(t^4 −10t^2 +5))/(5t^4 −10t^2 +1)) =((3t−t^2 )(5t^4 −10t^2 +1)+(t−3t^3 )(t^4 −10t^2 +5))/((1−3t^2 )(5t^4 −10t^2 +1))) =((−3t^7 −25t^3 −10t^2 +31t^5 +5t^4 +5t+1)/(−15t^6 +35t^4 −13t^2 +1)) soit ((3t^7 −31t^5 −5t^4 +25t^3 +10t^2 −5t+1)/(15t^6 −35t^4 +13t^2 −1))=2 30t^6 −70t^4 +26t^2 −2= 3t^7 −31t^5 −5t^4 +25t^3 +10t^2 −5t+1 3t^7 −30t^6 −31t^5 +65t^4 +25t^3 −16t^2 −5t+3=0 t_1 =−1,668557054 t_2 =1,1773794637 t_3 =20,76657908 x=−59 x_2 =49,65 x_3 =84,69 valeur retenue est x ≅49,65...
$$\begin{cases}{\mathrm{tan}\:\mathrm{a}+\mathrm{tan}\:\mathrm{b}=\frac{\mathrm{tan}\:\left(\mathrm{a}+\mathrm{b}\right)}{\mathrm{1}−\mathrm{tan}\:\mathrm{atan}\:\mathrm{b}}}\\{\mathrm{avec}\:\boldsymbol{\mathrm{t}}=\mathrm{tan}\:\boldsymbol{\mathrm{x}}\:\:\:\:\:\mathrm{tan}\:\mathrm{2x}=\frac{\mathrm{2t}}{\mathrm{2}−\mathrm{t}^{\mathrm{2}} }}\end{cases} \\ $$$$\mathrm{2}=\frac{\mathrm{tan}\:\mathrm{8}\boldsymbol{\mathrm{x}}}{\mathrm{1}−\mathrm{tan}\:\mathrm{3}\boldsymbol{\mathrm{x}}.\mathrm{tan}\:\mathrm{5}\boldsymbol{\mathrm{x}}} \\ $$$$\bullet\mathrm{tan}\:\mathrm{3x}=\frac{\mathrm{tan}\:\mathrm{2x}+\mathrm{tan}\:\mathrm{x}}{\mathrm{1}−\mathrm{tan}\:\mathrm{2x}.\mathrm{tan}\:\mathrm{x}}=\frac{\mathrm{t}\left(\mathrm{3}−\mathrm{t}^{\mathrm{2}} \right)}{\mathrm{1}−\mathrm{3t}^{\mathrm{2}} } \\ $$$$ \\ $$$$\:\:\:\mathrm{tan}\:\mathrm{3}\boldsymbol{\mathrm{x}}=\frac{\left(\mathrm{3}−\boldsymbol{\mathrm{t}}\right)\boldsymbol{\mathrm{t}}}{\left(\mathrm{1}−\mathrm{3}\boldsymbol{\mathrm{t}}^{\mathrm{2}} \right.}\:\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\bullet\mathrm{tan}\:\mathrm{5x}=\frac{\mathrm{tan}\:\mathrm{2x}+\mathrm{tan}\:\mathrm{3x}}{\mathrm{1}−\mathrm{tan}\:\mathrm{2x}.\mathrm{tan}\:\mathrm{3x}} \\ $$$$\:\:\:=\frac{\mathrm{2t}\left(\mathrm{1}−\mathrm{3t}^{\mathrm{2}} \right)+\mathrm{t}\left(\mathrm{1}−\mathrm{t}^{\mathrm{2}} \right)\left(\mathrm{3}−\mathrm{t}^{\mathrm{2}} \right)}{\left(\mathrm{1}−\mathrm{t}^{\mathrm{2}} \right)\left(\mathrm{1}−\mathrm{3t}^{\mathrm{2}} \right)−\mathrm{2t}^{\mathrm{2}} \left(\mathrm{3}−\mathrm{t}^{\mathrm{2}} \right)} \\ $$$$ \\ $$$$\:\:\:\:\:\mathrm{tan}\:\mathrm{5}\boldsymbol{\mathrm{x}}=\frac{\boldsymbol{\mathrm{t}}\left(\boldsymbol{\mathrm{t}}^{\mathrm{4}} −\mathrm{10}\boldsymbol{\mathrm{t}}^{\mathrm{2}} +\mathrm{5}\right)}{\mathrm{5}\boldsymbol{\mathrm{t}}^{\mathrm{4}} −\mathrm{10}\boldsymbol{\mathrm{t}}^{\mathrm{2}} +\mathrm{1}}\:\:\left(\mathrm{2}\right) \\ $$$$ \\ $$$$\mathrm{2}=\mathrm{tan}\:\mathrm{3}\boldsymbol{\mathrm{x}}+\mathrm{tan}\:\mathrm{5}\boldsymbol{\mathrm{x}}\:\:\:\:\Rightarrow \\ $$$$\mathrm{2}=\frac{\left(\mathrm{3}−\boldsymbol{\mathrm{t}}\right)\boldsymbol{\mathrm{t}}}{\mathrm{1}−\mathrm{3}\boldsymbol{\mathrm{t}}^{\mathrm{2}} }+\frac{\boldsymbol{\mathrm{t}}\left(\boldsymbol{\mathrm{t}}^{\mathrm{4}} −\mathrm{10}\boldsymbol{\mathrm{t}}^{\mathrm{2}} +\mathrm{5}\right)}{\mathrm{5}\boldsymbol{\mathrm{t}}^{\mathrm{4}} −\mathrm{10}\boldsymbol{\mathrm{t}}^{\mathrm{2}} +\mathrm{1}} \\ $$$$=\frac{\left.\mathrm{3}\boldsymbol{\mathrm{t}}−\boldsymbol{\mathrm{t}}^{\mathrm{2}} \right)\left(\mathrm{5}\boldsymbol{\mathrm{t}}^{\mathrm{4}} −\mathrm{10}\boldsymbol{\mathrm{t}}^{\mathrm{2}} +\mathrm{1}\right)+\left(\boldsymbol{\mathrm{t}}−\mathrm{3}\boldsymbol{\mathrm{t}}^{\mathrm{3}} \right)\left(\boldsymbol{\mathrm{t}}^{\mathrm{4}} −\mathrm{10}\boldsymbol{\mathrm{t}}^{\mathrm{2}} +\mathrm{5}\right)}{\left(\mathrm{1}−\mathrm{3}\boldsymbol{\mathrm{t}}^{\mathrm{2}} \right)\left(\mathrm{5}\boldsymbol{\mathrm{t}}^{\mathrm{4}} −\mathrm{10}\boldsymbol{\mathrm{t}}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$ \\ $$$$=\frac{−\mathrm{3}\boldsymbol{\mathrm{t}}^{\mathrm{7}} −\mathrm{25}\boldsymbol{\mathrm{t}}^{\mathrm{3}} −\mathrm{10}\boldsymbol{\mathrm{t}}^{\mathrm{2}} +\mathrm{31}\boldsymbol{\mathrm{t}}^{\mathrm{5}} +\mathrm{5}\boldsymbol{\mathrm{t}}^{\mathrm{4}} +\mathrm{5}\boldsymbol{\mathrm{t}}+\mathrm{1}}{−\mathrm{15}\boldsymbol{\mathrm{t}}^{\mathrm{6}} +\mathrm{35}\boldsymbol{\mathrm{t}}^{\mathrm{4}} −\mathrm{13}\boldsymbol{\mathrm{t}}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\boldsymbol{\mathrm{soit}} \\ $$$$\frac{\mathrm{3}\boldsymbol{\mathrm{t}}^{\mathrm{7}} −\mathrm{31}\boldsymbol{\mathrm{t}}^{\mathrm{5}} −\mathrm{5}\boldsymbol{\mathrm{t}}^{\mathrm{4}} +\mathrm{25}\boldsymbol{\mathrm{t}}^{\mathrm{3}} +\mathrm{10}\boldsymbol{\mathrm{t}}^{\mathrm{2}} −\mathrm{5}\boldsymbol{\mathrm{t}}+\mathrm{1}}{\mathrm{15}\boldsymbol{\mathrm{t}}^{\mathrm{6}} −\mathrm{35}\boldsymbol{\mathrm{t}}^{\mathrm{4}} +\mathrm{13}\boldsymbol{\mathrm{t}}^{\mathrm{2}} −\mathrm{1}}=\mathrm{2} \\ $$$$\mathrm{30}\boldsymbol{\mathrm{t}}^{\mathrm{6}} −\mathrm{70}\boldsymbol{\mathrm{t}}^{\mathrm{4}} +\mathrm{26}\boldsymbol{\mathrm{t}}^{\mathrm{2}} −\mathrm{2}= \\ $$$$\mathrm{3}\boldsymbol{\mathrm{t}}^{\mathrm{7}} −\mathrm{31}\boldsymbol{\mathrm{t}}^{\mathrm{5}} −\mathrm{5}\boldsymbol{\mathrm{t}}^{\mathrm{4}} +\mathrm{25}\boldsymbol{\mathrm{t}}^{\mathrm{3}} +\mathrm{10}\boldsymbol{\mathrm{t}}^{\mathrm{2}} −\mathrm{5}\boldsymbol{\mathrm{t}}+\mathrm{1} \\ $$$$ \\ $$$$\mathrm{3}\boldsymbol{\mathrm{t}}^{\mathrm{7}} −\mathrm{30}\boldsymbol{\mathrm{t}}^{\mathrm{6}} −\mathrm{31}\boldsymbol{\mathrm{t}}^{\mathrm{5}} +\mathrm{65}\boldsymbol{\mathrm{t}}^{\mathrm{4}} +\mathrm{25}\boldsymbol{\mathrm{t}}^{\mathrm{3}} −\mathrm{16}\boldsymbol{\mathrm{t}}^{\mathrm{2}} −\mathrm{5}\boldsymbol{\mathrm{t}}+\mathrm{3}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{t}_{\mathrm{1}} =−\mathrm{1},\mathrm{668557054} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{t}_{\mathrm{2}} =\mathrm{1},\mathrm{1773794637} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{t}_{\mathrm{3}} =\mathrm{20},\mathrm{76657908} \\ $$$$\:\mathrm{x}=−\mathrm{59}\:\:\:\:\:\boldsymbol{\mathrm{x}}_{\mathrm{2}} =\mathrm{49},\mathrm{65}\:\:\:\:\:\:\mathrm{x}_{\mathrm{3}} =\mathrm{84},\mathrm{69} \\ $$$$\mathrm{valeur}\:\mathrm{retenue} \\ $$$$ \\ $$$$\mathrm{est}\:\boldsymbol{\mathrm{x}}\:\:\cong\mathrm{49},\mathrm{65}…\:\:\:\: \\ $$$$ \\ $$
Commented by a.lgnaoui last updated on 21/Jul/24
Commented by hardmath last updated on 24/Jul/24
thank you dear professors
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professors} \\ $$
Answered by Frix last updated on 22/Jul/24
tan 3x =(((3−tan^2 x)tan x)/(1−3tan^2 x)) tan 5x =(((5−10tan^2 x +4tan^4 x))/(1−10tan^2 x +5tan^4 x)) tan 3x +tan 5x =2 ∧ t=tan x ⇒ t^7 −((15t^6 )/4)−7t^5 +((35t^4 )/4)+7t^3 −((13t^2 )/4)−t+(1/4)=0 This can only be approximated.
$$\mathrm{tan}\:\mathrm{3}{x}\:=\frac{\left(\mathrm{3}−\mathrm{tan}^{\mathrm{2}} \:{x}\right)\mathrm{tan}\:{x}}{\mathrm{1}−\mathrm{3tan}^{\mathrm{2}} \:{x}} \\ $$$$\mathrm{tan}\:\mathrm{5}{x}\:=\frac{\left(\mathrm{5}−\mathrm{10tan}^{\mathrm{2}} \:{x}\:+\mathrm{4tan}^{\mathrm{4}} \:{x}\right)}{\mathrm{1}−\mathrm{10tan}^{\mathrm{2}} \:{x}\:+\mathrm{5tan}^{\mathrm{4}} \:{x}} \\ $$$$\mathrm{tan}\:\mathrm{3}{x}\:+\mathrm{tan}\:\mathrm{5}{x}\:=\mathrm{2}\:\wedge\:{t}=\mathrm{tan}\:{x} \\ $$$$\Rightarrow \\ $$$${t}^{\mathrm{7}} −\frac{\mathrm{15}{t}^{\mathrm{6}} }{\mathrm{4}}−\mathrm{7}{t}^{\mathrm{5}} +\frac{\mathrm{35}{t}^{\mathrm{4}} }{\mathrm{4}}+\mathrm{7}{t}^{\mathrm{3}} −\frac{\mathrm{13}{t}^{\mathrm{2}} }{\mathrm{4}}−{t}+\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{0} \\ $$$$\mathrm{This}\:\mathrm{can}\:\mathrm{only}\:\mathrm{be}\:\mathrm{approximated}. \\ $$

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