tan(3x) + tan(5x) = 2 Find: x = ?


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Question Number 209735 by hardmath last updated on 19/Jul/24

$\tan (3 x) + \tan (5 x) = 2$ $Find : x = ?$
	Answered by a.lgnaoui last updated on 20/Jul/24

	$voir reponse en bas$ $(a l exeption d^{'} ereur de calcul)$
	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
	${ ((tan a+tan b=((tan (a+b))/(1−tan atan b)))),((avec t=tan x tan 2x=((2t)/(2−t^2 )))) :} 2=((tan 8x)/(1−tan 3x.tan 5x)) •tan 3x=((tan 2x+tan x)/(1−tan 2x.tan x))=((t(3−t^2 ))/(1−3t^2 )) tan 3x=(((3−t)t)/((1−3t^2 )) (1) •tan 5x=((tan 2x+tan 3x)/(1−tan 2x.tan 3x)) =((2t(1−3t^2 )+t(1−t^2 )(3−t^2 ))/((1−t^2 )(1−3t^2 )−2t^2 (3−t^2 ))) tan 5x=((t(t^4 −10t^2 +5))/(5t^4 −10t^2 +1)) (2) 2=tan 3x+tan 5x ⇒ 2=(((3−t)t)/(1−3t^2 ))+((t(t^4 −10t^2 +5))/(5t^4 −10t^2 +1)) =((3t−t^2 )(5t^4 −10t^2 +1)+(t−3t^3 )(t^4 −10t^2 +5))/((1−3t^2 )(5t^4 −10t^2 +1))) =((−3t^7 −25t^3 −10t^2 +31t^5 +5t^4 +5t+1)/(−15t^6 +35t^4 −13t^2 +1)) soit ((3t^7 −31t^5 −5t^4 +25t^3 +10t^2 −5t+1)/(15t^6 −35t^4 +13t^2 −1))=2 30t^6 −70t^4 +26t^2 −2= 3t^7 −31t^5 −5t^4 +25t^3 +10t^2 −5t+1 3t^7 −30t^6 −31t^5 +65t^4 +25t^3 −16t^2 −5t+3=0 t_1 =−1,668557054 t_2 =1,1773794637 t_3 =20,76657908 x=−59 x_2 =49,65 x_3 =84,69 valeur retenue est x ≅49,65...$
	${\begin{cases} \tan a + \tan b = \frac{\tan (a + b)}{1 - \tan atan b} \\ avec t = \tan x \tan 2 x = \frac{2 t}{2 - t^{2}} \end{cases}$ $2 = \frac{\tan 8 x}{1 - \tan 3 x . \tan 5 x}$ $∙ \tan 3 x = \frac{\tan 2 x + \tan x}{1 - \tan 2 x . \tan x} = \frac{t (3 - t^{2})}{1 - {3 t}^{2}}$ $\tan 3 x = \frac{(3 - t) t}{(1 - 3 t^{2}} (1)$ $∙ \tan 5 x = \frac{\tan 2 x + \tan 3 x}{1 - \tan 2 x . \tan 3 x}$ $= \frac{2 t (1 - {3 t}^{2}) + t (1 - t^{2}) (3 - t^{2})}{(1 - t^{2}) (1 - {3 t}^{2}) - {2 t}^{2} (3 - t^{2})}$ $\tan 5 x = \frac{t (t^{4} - 10 t^{2} + 5)}{5 t^{4} - 10 t^{2} + 1} (2)$ $2 = \tan 3 x + \tan 5 x \Rightarrow$ $2 = \frac{(3 - t) t}{1 - 3 t^{2}} + \frac{t (t^{4} - 10 t^{2} + 5)}{5 t^{4} - 10 t^{2} + 1}$ $= \frac{3 t - t^{2}) (5 t^{4} - 10 t^{2} + 1) + (t - 3 t^{3}) (t^{4} - 10 t^{2} + 5)}{(1 - 3 t^{2}) (5 t^{4} - 10 t^{2} + 1)}$ $= \frac{- 3 t^{7} - 25 t^{3} - 10 t^{2} + 31 t^{5} + 5 t^{4} + 5 t + 1}{- 15 t^{6} + 35 t^{4} - 13 t^{2} + 1}$ $soit$ $\frac{3 t^{7} - 31 t^{5} - 5 t^{4} + 25 t^{3} + 10 t^{2} - 5 t + 1}{15 t^{6} - 35 t^{4} + 13 t^{2} - 1} = 2$ $30 t^{6} - 70 t^{4} + 26 t^{2} - 2 =$ $3 t^{7} - 31 t^{5} - 5 t^{4} + 25 t^{3} + 10 t^{2} - 5 t + 1$ $3 t^{7} - 30 t^{6} - 31 t^{5} + 65 t^{4} + 25 t^{3} - 16 t^{2} - 5 t + 3 = 0$ $t_{1} = - 1, 668557054$ $t_{2} = 1, 1773794637$ $t_{3} = 20, 76657908$ $x = - 59 x_{2} = 49, 65 x_{3} = 84, 69$ $valeur retenue$ $est x ≅ 49, 65 . . .$
	Commented by a.lgnaoui last updated on 21/Jul/24

	Commented by hardmath last updated on 24/Jul/24

	$thank you dear professors$
	Answered by Frix last updated on 22/Jul/24

	$\tan 3 x = \frac{(3 - \tan^{2} x) \tan x}{1 - {3 \tan}^{2} x}$ $\tan 5 x = \frac{(5 - {10 \tan}^{2} x + {4 \tan}^{4} x)}{1 - {10 \tan}^{2} x + {5 \tan}^{4} x}$ $\tan 3 x + \tan 5 x = 2 \land t = \tan x$ $\Rightarrow$ $t^{7} - \frac{15 t^{6}}{4} - 7 t^{5} + \frac{35 t^{4}}{4} + 7 t^{3} - \frac{13 t^{2}}{4} - t + \frac{1}{4} = 0$ $This can only be approximated .$
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