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Question Number 40100 by maxmathsup by imad last updated on 15/Jul/18

solve arctan(2x) +arctan(3x)=(π/4)

$${solve}\:\:\:{arctan}\left(\mathrm{2}{x}\right)\:+{arctan}\left(\mathrm{3}{x}\right)=\frac{\pi}{\mathrm{4}} \\ $$

Commented by math khazana by abdo last updated on 26/Jul/18

⇒tan{arctan(2x)+arctan(3x)}=1 ⇒ ((2x +3x)/(1−2x.3x)) =1 ⇒ ((5x)/(1−6x^2 )) =1 ⇒ ((5x)/(1−6x^2 )) −1=0 ⇒((5x−1+6x^2 )/(1−6x^2 )) =0 ⇒ 6x^2 +5x−1=0 and x^2 ≠ (1/6) ⇒x ≠+^− (1/(√6)) Δ =25−4(6)(−1) =25+24=49 x_1 =((−5 +7)/(12)) =(2/(12)) =(1/6) x_2 =((−5−7)/(12)) =−1

$$\Rightarrow{tan}\left\{{arctan}\left(\mathrm{2}{x}\right)+{arctan}\left(\mathrm{3}{x}\right)\right\}=\mathrm{1}\:\Rightarrow \\ $$$$\frac{\mathrm{2}{x}\:+\mathrm{3}{x}}{\mathrm{1}−\mathrm{2}{x}.\mathrm{3}{x}}\:=\mathrm{1}\:\Rightarrow\:\frac{\mathrm{5}{x}}{\mathrm{1}−\mathrm{6}{x}^{\mathrm{2}} }\:=\mathrm{1}\:\Rightarrow \\ $$$$\frac{\mathrm{5}{x}}{\mathrm{1}−\mathrm{6}{x}^{\mathrm{2}} }\:−\mathrm{1}=\mathrm{0}\:\Rightarrow\frac{\mathrm{5}{x}−\mathrm{1}+\mathrm{6}{x}^{\mathrm{2}} }{\mathrm{1}−\mathrm{6}{x}^{\mathrm{2}} }\:=\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{6}{x}^{\mathrm{2}} \:+\mathrm{5}{x}−\mathrm{1}=\mathrm{0}\:{and}\:{x}^{\mathrm{2}} \neq\:\frac{\mathrm{1}}{\mathrm{6}}\:\Rightarrow{x}\:\neq\overset{−} {+}\frac{\mathrm{1}}{\sqrt{\mathrm{6}}} \\ $$$$\Delta\:=\mathrm{25}−\mathrm{4}\left(\mathrm{6}\right)\left(−\mathrm{1}\right)\:=\mathrm{25}+\mathrm{24}=\mathrm{49} \\ $$$${x}_{\mathrm{1}} =\frac{−\mathrm{5}\:+\mathrm{7}}{\mathrm{12}}\:=\frac{\mathrm{2}}{\mathrm{12}}\:=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$${x}_{\mathrm{2}} =\frac{−\mathrm{5}−\mathrm{7}}{\mathrm{12}}\:=−\mathrm{1}\:\: \\ $$

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