Find all sum of all x interval [ 0, 2π ] such that 3cot^2 x + 8 cot x + 3 = 0


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Question Number 119373 by bobhans last updated on 24/Oct/20

$F i n d a l l s u m o f a l l x i n t e r v a l$ $[0, 2 π] s u c h t h a t 3 \cot^{2} x + 8 \cot x + 3 = 0$
Commented by benjo_mathlover last updated on 24/Oct/20

$c o n s i d e r t h e q u a d r a t i c e q u a t i o n$ $3 q^{2} + 8 q + 3 = 0, w i t h t h e r o o t s a r e q_{1} = \frac{- 8 + 2 \sqrt{7}}{6}$ $a n d q_{2} = \frac{- 8 - 2 \sqrt{7}}{6} . b o t h r o o t s a r e r e a l a n d$ $t h e i r p r o d u c t e q u a l t o - 1 . B e c a u s e y = \cot x i s a b i j e c t i o n$ $f r o m t h e i n t e r v a l (0, π) t o t h e r e a l n u m b e r, t h e r e$ $i s a u n i q u e p a i r o f n u m b e r s x_{1.1} a n d x_{2.1}$ $w i t h 0 < x_{1.1}, x_{2.1} < π s u c h t h a t \cot x_{1.1} = q_{1}$ $a n d \cot x_{2.1} = q_{2} . B e c a u s e q_{1} a n d q_{2} a r e$ $n e g a t i v e, \frac{π}{2} < x_{1.1}, x_{2.1} < π a n d s o$ $π < x_{1.1}, x_{2.1} < 2 π . S i n c e \tan x \cot x = 1$ $b o t h \tan x a n d \cot x h a v e p e r i o d π, i t f o l l o w s t h a t$ $1 = \tan x \cot x = \cot x \cot (\frac{π}{2} - x) = \cot x \cot (\frac{3 π}{2} - x)$ $1 = \cot x_{1.1} . \cot x_{2.1} t h e n x_{1.1} + x_{2.1} = \frac{3 π}{2}$ $l i k e w i s e i n i n t e r v a l (π, 2 π) t h e r e i s u n i q u e$ $p a i r n u m b e r x_{1.2} a n d x_{2.2} s a t i s f y i n g t h e$ $c o n d i t i o n s o f t h e p r o b l e m w i t h x_{1.2} + x_{2.2} = \frac{7 π}{2}$ $T h e r e f o r e x_{1.1} + x_{2.1} + x_{1.2} + x_{2.2} = \frac{3 π}{2} + \frac{7 π}{2} = 5 π$
	Answered by TANMAY PANACEA last updated on 24/Oct/20

	$t a n x = a$ $\frac{3}{a^{2}} + \frac{8}{a} + 3 = 0$ $3 + 8 a + 3 a^{2} = 0$ $a = \frac{- 8 \pm \sqrt{8^{2} - 4 \times 3 \times 3}}{2 \times 3} = \frac{- 8 \pm 2 \sqrt{7}}{2 \times 3} = \frac{- 4 \pm \sqrt{7}}{3}$ ${t a n x}_{1} = \frac{- 4 + \sqrt{7}}{3} = t a n α {t a n x}_{2} = \frac{- 4 - \sqrt{7}}{3} = t a n β$ $t a n (α + β) = \frac{t a n α + t a n β}{1 - t a n α t a n β} = \frac{\frac{- 8}{3}}{1 - (\frac{16 - 7}{9})} = t a n \frac{π}{2}$ $(α + β) = \frac{π}{2}$ $t a n (α + β) = t a n \frac{π}{2} = t a n (π + \frac{π}{2}) = t a n \frac{3 π}{2}$ $t a n α = t a n (π + α)$ $t a n β = t a n (π + β)$ $α + (α + π) + β + (β + π) + (α + β) + (π + α + β)$ $= 4 (α + β) + 3 π$ $= 4 (\frac{π}{2}) + 3 π = 5 π$
	Commented by 1549442205PVT last updated on 24/Oct/20

	$Sir mistake at place \sqrt{8^{2} - 4 \times 3 \times 3}$ $= \sqrt{28} = 2 \sqrt{7}$
	Commented by TANMAY PANACEA last updated on 24/Oct/20

	$y e s s i r c o r r e c t e d$
	Answered by MJS_new last updated on 24/Oct/20

	${3 \cot}^{2} x + 8 \cot x + 3 = 0$ $\frac{{3 \cos}^{2} x}{\sin^{2} x} + \frac{8 \cos x}{\sin x} + 3 = 0$ $\frac{{3 \cos}^{2} + 8 \cos x \sin x + {3 \sin}^{2} x}{\sin^{2} x} = 0$ $3 + 8 \cos x \sin x = 0$ $\cos x \sin x = - \frac{3}{8}$ $\frac{1}{2} \sin 2 x = - \frac{3}{8}$ $\sin 2 x = - \frac{3}{4}$ $x = n π - \frac{1}{2} \arcsin \frac{3}{4} \lor x = n π + \frac{π}{2} + \frac{1}{2} \arcsin \frac{3}{4}$ $let α = \frac{1}{2} \arcsin \frac{3}{4}$ $for 0 ⩽ x < 2 π we get$ $x = π - α \lor x = 2 π - α \lor x = \frac{π}{2} + α \lor x = \frac{3 π}{2} + α$ $\Rightarrow sum of these is 5 π$
	Answered by 1549442205PVT last updated on 24/Oct/20
	$3cot^2 x + 8 cot x + 3 = 0 Δ′=4^2 −9=7⇒cotx=((−4±(√7))/3) ⇒tanx=(3/(−4±(√7)))=((3(−4∓(√7)))/9)=((−4∓(√7))/3) Put α=tan^(−1) (((−4+(√7))/3)),β=tan^(−1) (((−4−(√7))/3)) α≈−24°17′42”,β≈−65°42′17” Since α,β∈[0,2π] and tanx=tan(x+kπ) we get α∈{155°42′17”,335°42′17”} β∈{114°17′43”,294°17′43”} ⇒α_1 +α_2 +β_1 +β_2 =900° Thus,Sum of all the angles satisfying given equation equal to 900°$
	$3 \cot^{2} x + 8 \cot x + 3 = 0$ $Δ^{'} = 4^{2} - 9 = 7 \Rightarrow cotx = \frac{- 4 \pm \sqrt{7}}{3}$ $\Rightarrow tanx = \frac{3}{- 4 \pm \sqrt{7}} = \frac{3 (- 4 \mp \sqrt{7})}{9} = \frac{- 4 \mp \sqrt{7}}{3}$ $Put α = \tan^{- 1} (\frac{- 4 + \sqrt{7}}{3}), β = \tan^{- 1} (\frac{- 4 - \sqrt{7}}{3})$ $α \approx - 24 ° 17^{'} 42^{″}, β \approx - 65 ° 42^{'} 17^{″}$ $Since α, β \in [0, 2 π] and tanx = \tan (x + k π)$ $we get α \in {155 ° 42^{'} 17^{″}, 335 ° 42^{'} 17^{″}}$ $β \in {114 ° 17^{'} 43^{″}, 294 ° 17^{'} 43^{″}}$ $\Rightarrow α_{1} + α_{2} + β_{1} + β_{2} = 900 °$ $Thus, Sum of all the angles satisfying$ $given equation equal to 900 °$
	Answered by Bird last updated on 24/Oct/20

	$3 {c o t a n}^{2} x + 8 c o t a n x + 3 = 0 \Rightarrow$ $\frac{3}{{t a n}^{2} x} + \frac{8}{t a n x} + 3 = 0 \Rightarrow$ $3 + 8 t a n x + 3 {t a n}^{2} x = 0 \Rightarrow$ $3 {t a n}^{2} x + 8 t a n x + 3 = 0$ $t a n x = u \Rightarrow 3 u^{2} + 8 u + 3 = 0$ $Δ^{^{'}} = 4^{2} - 9 = 16 - 9 = 7 \Rightarrow$ $u_{1} = \frac{- 4 + \sqrt{7}}{3} a n d u_{2} = \frac{- 4 - \sqrt{7}}{3}$ $t a n x = \frac{- 4 + \sqrt{7}}{3} \Rightarrow x = a r c t a n (\frac{- 4 + \sqrt{7}}{3}) + n π$ $t a n x = \frac{- 4 - \sqrt{7}}{3} \Rightarrow x = - a r c t a n (\frac{4 + \sqrt{7}}{3}) + n π$ $(n \in Z)$
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