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

(2tanx−5)tanx+(2cotx−5)cotx−8=0

(2tanx5)tanx+(2cotx5)cotx8=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...

(2a5)a+(2a5)1a8=0(2a25a)+(25aa)1a8=02a25a+25aa28=02a45a3+25a8a2=02a45a38a25a+2=02a25a85a+2a2=02(a2+1a2)5(a+1a)8=0k=a+1aa2+1a2=k222(k22)5k8=02k245k8=02k25k12=02k28k+3k12=02k(k4)+3(k4)=0(k4)(2k+3)=0k=4k=32a+1a=4a24a+1=0a=4±1642=4±232=2±3a+1a=32a2+1a=322a2+3a+2=0a=3±94.2.22=3±i72notfeasiblesolutionsoa=tanx=2+3x=tan1(2+3)tanx=(2+3)=tan1{(2+3)}1+13113=3+131=4+232=2+3sotanx=tan(Π4+Π6)=tan(5Π12)x=nΠ+5Π12....whentanx=(2+3)tanx=tan(5Π12)=tan(Π5Π12)=tan7Π12x=nΠ+7Π12plscheck...

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

thksyou.butIfound:x=Π12+nΠorx=5Π12+nΠ.Isittrue?bylog10b

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