(2tanx−5)tanx+(2cotx−5)cotx−8=0


Question and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 42170 by lucha116 last updated on 19/Aug/18

$(2 t a n x - 5) t a n x + (2 c o t x - 5) c o t x - 8 = 0$
	Answered by tanmay.chaudhury50@gmail.com last updated on 19/Aug/18
	$(2a−5)a+((2/a)−5)(1/a)−8=0 (2a^2 −5a)+(((2−5a)/a))(1/a)−8=0 2a^2 −5a+((2−5a)/a^2 )−8=0 2a^4 −5a^3 +2−5a−8a^2 =0 2a^4 −5a^3 −8a^2 −5a+2=0 2a^2 −5a−8−(5/a)+(2/a^2 )=0 2(a^2 +(1/a^2 ))−5(a+(1/a))−8=0 k=a+(1/a) a^2 +(1/a^2 )=k^2 −2 2(k^2 −2)−5k−8=0 2k^2 −4−5k−8=0 2k^2 −5k−12=0 2k^2 −8k+3k−12=0 2k(k−4)+3(k−4)=0 (k−4)(2k+3)=0 k=4 k=((−3)/2) a+(1/a)=4 a^2 −4a+1=0 a=((4±(√(16−4)) )/2)=((4±2(√(3 )))/2)=2±(√3) a+(1/a)=((−3)/2) ((a^2 +1)/a)=((−3)/2) 2a^2 +3a+2=0 a=((−3±(√(9−4.2.2)))/2)=((−3±i(√7))/2) not feasible solution so a=tanx=2+(√3) x=tan^(−1) (2+(√3) ) tanx=−(2+(√3) )=tan^(−1) {−(2+(√3) )} ((1+(1/(√3)))/(1−(1/(√3))))=(((√3) +1)/((√3) −1)) =((4+2(√3))/2)=2+(√3) so tanx=tan((Π/4)+(Π/6))=tan(((5Π)/(12))) x=nΠ+((5Π)/(12)).... when tanx=−(2+(√(3))) tanx=−tan(((5Π)/(12)))=tan(Π−((5Π)/(12)))=tan((7Π)/(12)) x=nΠ+((7Π)/(12)) pls check...$
	$(2 a - 5) a + (\frac{2}{a} - 5) \frac{1}{a} - 8 = 0$ $(2 a^{2} - 5 a) + (\frac{2 - 5 a}{a}) \frac{1}{a} - 8 = 0$ $2 a^{2} - 5 a + \frac{2 - 5 a}{a^{2}} - 8 = 0$ $2 a^{4} - 5 a^{3} + 2 - 5 a - 8 a^{2} = 0$ $2 a^{4} - 5 a^{3} - 8 a^{2} - 5 a + 2 = 0$ $2 a^{2} - 5 a - 8 - \frac{5}{a} + \frac{2}{a^{2}} = 0$ $2 (a^{2} + \frac{1}{a^{2}}) - 5 (a + \frac{1}{a}) - 8 = 0$ $k = a + \frac{1}{a} a^{2} + \frac{1}{a^{2}} = k^{2} - 2$ $2 (k^{2} - 2) - 5 k - 8 = 0$ $2 k^{2} - 4 - 5 k - 8 = 0$ $2 k^{2} - 5 k - 12 = 0$ $2 k^{2} - 8 k + 3 k - 12 = 0$ $2 k (k - 4) + 3 (k - 4) = 0$ $(k - 4) (2 k + 3) = 0$ $k = 4 k = \frac{- 3}{2}$ $a + \frac{1}{a} = 4$ $a^{2} - 4 a + 1 = 0$ $a = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm 2 \sqrt{3}}{2} = 2 \pm \sqrt{3}$ $a + \frac{1}{a} = \frac{- 3}{2}$ $\frac{a^{2} + 1}{a} = \frac{- 3}{2}$ $2 a^{2} + 3 a + 2 = 0$ $a = \frac{- 3 \pm \sqrt{9 - 4.2.2}}{2} = \frac{- 3 \pm i \sqrt{7}}{2} n o t f e a s i b l e s o l u t i o n$ $s o a = t a n x = 2 + \sqrt{3} x = {t a n}^{- 1} (2 + \sqrt{3})$ $t a n x = - (2 + \sqrt{3}) = {t a n}^{- 1} {- (2 + \sqrt{3})}$ $\frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} = \frac{4 + 2 \sqrt{3}}{2} = 2 + \sqrt{3}$ $s o t a n x = t a n (\frac{Π}{4} + \frac{Π}{6}) = t a n (\frac{5 Π}{12})$ $x = n Π + \frac{5 Π}{12} . . . .$ $w h e n t a n x = - (2 + \sqrt{3)}$ $t a n x = - t a n (\frac{5 Π}{12}) = t a n (Π - \frac{5 Π}{12}) = t a n \frac{7 Π}{12}$ $x = n Π + \frac{7 Π}{12}$ $p l s c h e c k . . .$
	Commented by lucha116 last updated on 20/Aug/18

	$t h k s y o u . b u t I f o u n d : x = \frac{Π}{12} + n Π o r x = \frac{5 Π}{12} + n Π .$ $I s i t t r u e ?$ $b y \log_{10}$ $b$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum