Find-all-sum-of-all-x-interval-0-2pi-such-that-3cot-2-x-8-cot-x-3-0-

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

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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