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Question Number 42170 by lucha116 last updated on 19/Aug/18

(2tanx−5)tanx+(2cotx−5)cotx−8=0

$$\left(\mathrm{2}{tanx}−\mathrm{5}\right){tanx}+\left(\mathrm{2}{cotx}−\mathrm{5}\right){cotx}−\mathrm{8}=\mathrm{0} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 19/Aug/18

(2a−5)a+((2/a)−5)(1/a)−8=0 (2a^2 −5a)+(((2−5a)/a))(1/a)−8=0 2a^2 −5a+((2−5a)/a^2 )−8=0 2a^4 −5a^3 +2−5a−8a^2 =0 2a^4 −5a^3 −8a^2 −5a+2=0 2a^2 −5a−8−(5/a)+(2/a^2 )=0 2(a^2 +(1/a^2 ))−5(a+(1/a))−8=0 k=a+(1/a) a^2 +(1/a^2 )=k^2 −2 2(k^2 −2)−5k−8=0 2k^2 −4−5k−8=0 2k^2 −5k−12=0 2k^2 −8k+3k−12=0 2k(k−4)+3(k−4)=0 (k−4)(2k+3)=0 k=4 k=((−3)/2) a+(1/a)=4 a^2 −4a+1=0 a=((4±(√(16−4)) )/2)=((4±2(√(3 )))/2)=2±(√3) a+(1/a)=((−3)/2) ((a^2 +1)/a)=((−3)/2) 2a^2 +3a+2=0 a=((−3±(√(9−4.2.2)))/2)=((−3±i(√7))/2) not feasible solution so a=tanx=2+(√3) x=tan^(−1) (2+(√3) ) tanx=−(2+(√3) )=tan^(−1) {−(2+(√3) )} ((1+(1/(√3)))/(1−(1/(√3))))=(((√3) +1)/((√3) −1)) =((4+2(√3))/2)=2+(√3) so tanx=tan((Π/4)+(Π/6))=tan(((5Π)/(12))) x=nΠ+((5Π)/(12)).... when tanx=−(2+(√(3))) tanx=−tan(((5Π)/(12)))=tan(Π−((5Π)/(12)))=tan((7Π)/(12)) x=nΠ+((7Π)/(12)) pls check...

$$\left(\mathrm{2}{a}−\mathrm{5}\right){a}+\left(\frac{\mathrm{2}}{{a}}−\mathrm{5}\right)\frac{\mathrm{1}}{{a}}−\mathrm{8}=\mathrm{0} \\ $$$$\left(\mathrm{2}{a}^{\mathrm{2}} −\mathrm{5}{a}\right)+\left(\frac{\mathrm{2}−\mathrm{5}{a}}{{a}}\right)\frac{\mathrm{1}}{{a}}−\mathrm{8}=\mathrm{0} \\ $$$$\mathrm{2}{a}^{\mathrm{2}} −\mathrm{5}{a}+\frac{\mathrm{2}−\mathrm{5}{a}}{{a}^{\mathrm{2}} }−\mathrm{8}=\mathrm{0} \\ $$$$\mathrm{2}{a}^{\mathrm{4}} −\mathrm{5}{a}^{\mathrm{3}} +\mathrm{2}−\mathrm{5}{a}−\mathrm{8}{a}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{2}{a}^{\mathrm{4}} −\mathrm{5}{a}^{\mathrm{3}} −\mathrm{8}{a}^{\mathrm{2}} −\mathrm{5}{a}+\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{2}{a}^{\mathrm{2}} −\mathrm{5}{a}−\mathrm{8}−\frac{\mathrm{5}}{{a}}+\frac{\mathrm{2}}{{a}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\mathrm{2}\left({a}^{\mathrm{2}} +\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\right)−\mathrm{5}\left({a}+\frac{\mathrm{1}}{{a}}\right)−\mathrm{8}=\mathrm{0} \\ $$$${k}={a}+\frac{\mathrm{1}}{{a}}\:\:{a}^{\mathrm{2}} +\frac{\mathrm{1}}{{a}^{\mathrm{2}} }={k}^{\mathrm{2}} −\mathrm{2} \\ $$$$\mathrm{2}\left({k}^{\mathrm{2}} −\mathrm{2}\right)−\mathrm{5}{k}−\mathrm{8}=\mathrm{0} \\ $$$$\mathrm{2}{k}^{\mathrm{2}} −\mathrm{4}−\mathrm{5}{k}−\mathrm{8}=\mathrm{0} \\ $$$$\mathrm{2}{k}^{\mathrm{2}} −\mathrm{5}{k}−\mathrm{12}=\mathrm{0} \\ $$$$\mathrm{2}{k}^{\mathrm{2}} −\mathrm{8}{k}+\mathrm{3}{k}−\mathrm{12}=\mathrm{0} \\ $$$$\mathrm{2}{k}\left({k}−\mathrm{4}\right)+\mathrm{3}\left({k}−\mathrm{4}\right)=\mathrm{0} \\ $$$$\left({k}−\mathrm{4}\right)\left(\mathrm{2}{k}+\mathrm{3}\right)=\mathrm{0} \\ $$$${k}=\mathrm{4}\:\:\:\:{k}=\frac{−\mathrm{3}}{\mathrm{2}} \\ $$$${a}+\frac{\mathrm{1}}{{a}}=\mathrm{4} \\ $$$${a}^{\mathrm{2}} −\mathrm{4}{a}+\mathrm{1}=\mathrm{0} \\ $$$${a}=\frac{\mathrm{4}\pm\sqrt{\mathrm{16}−\mathrm{4}}\:}{\mathrm{2}}=\frac{\mathrm{4}\pm\mathrm{2}\sqrt{\mathrm{3}\:}}{\mathrm{2}}=\mathrm{2}\pm\sqrt{\mathrm{3}} \\ $$$${a}+\frac{\mathrm{1}}{{a}}=\frac{−\mathrm{3}}{\mathrm{2}} \\ $$$$\frac{{a}^{\mathrm{2}} +\mathrm{1}}{{a}}=\frac{−\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{2}{a}^{\mathrm{2}} +\mathrm{3}{a}+\mathrm{2}=\mathrm{0} \\ $$$${a}=\frac{−\mathrm{3}\pm\sqrt{\mathrm{9}−\mathrm{4}.\mathrm{2}.\mathrm{2}}}{\mathrm{2}}=\frac{−\mathrm{3}\pm{i}\sqrt{\mathrm{7}}}{\mathrm{2}}\:{not}\:{feasible}\:{solution} \\ $$$${so}\:{a}={tanx}=\mathrm{2}+\sqrt{\mathrm{3}}\:\:{x}={tan}^{−\mathrm{1}} \left(\mathrm{2}+\sqrt{\mathrm{3}}\:\right) \\ $$$${tanx}=−\left(\mathrm{2}+\sqrt{\mathrm{3}}\:\right)={tan}^{−\mathrm{1}} \left\{−\left(\mathrm{2}+\sqrt{\mathrm{3}}\:\right)\right\} \\ $$$$\frac{\mathrm{1}+\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}}{\mathrm{1}−\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}}=\frac{\sqrt{\mathrm{3}}\:+\mathrm{1}}{\sqrt{\mathrm{3}}\:−\mathrm{1}}\:=\frac{\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{2}}=\mathrm{2}+\sqrt{\mathrm{3}} \\ $$$${so}\:{tanx}={tan}\left(\frac{\Pi}{\mathrm{4}}+\frac{\Pi}{\mathrm{6}}\right)={tan}\left(\frac{\mathrm{5}\Pi}{\mathrm{12}}\right) \\ $$$${x}={n}\Pi+\frac{\mathrm{5}\Pi}{\mathrm{12}}.... \\ $$$${when}\:{tanx}=−\left(\mathrm{2}+\sqrt{\left.\mathrm{3}\right)}\right. \\ $$$$\:\:{tanx}=−{tan}\left(\frac{\mathrm{5}\Pi}{\mathrm{12}}\right)={tan}\left(\Pi−\frac{\mathrm{5}\Pi}{\mathrm{12}}\right)={tan}\frac{\mathrm{7}\Pi}{\mathrm{12}} \\ $$$${x}={n}\Pi+\frac{\mathrm{7}\Pi}{\mathrm{12}} \\ $$$${pls}\:{check}... \\ $$$$ \\ $$

Commented by lucha116 last updated on 20/Aug/18

thks you. but I found: x=(Π/(12))+nΠ or x=((5Π)/(12))+nΠ. Is it true? bylog_(10) b

$${thks}\:{you}.\:{but}\:{I}\:{found}:\:{x}=\frac{\Pi}{\mathrm{12}}+{n}\Pi\:{or}\:{x}=\frac{\mathrm{5}\Pi}{\mathrm{12}}+{n}\Pi.\: \\ $$$${Is}\:{it}\:{true}? \\ $$$${by}\mathrm{log}_{\mathrm{10}} \\ $$$${b} \\ $$

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