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

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

$${Find}\:{all}\:{sum}\:{of}\:{all}\:{x}\:{interval} \\ $$$$\left[\:\mathrm{0},\:\mathrm{2}\pi\:\right]\:{such}\:{that}\:\mathrm{3cot}\:^{\mathrm{2}} {x}\:+\:\mathrm{8}\:\mathrm{cot}\:{x}\:+\:\mathrm{3}\:=\:\mathrm{0} \\ $$

Commented by benjo_mathlover last updated on 24/Oct/20

consider the quadratic equation 3q^2 +8q+3 = 0 , with the roots are q_1 =((−8+2(√7))/6) and q_2 =((−8−2(√7))/6). both roots are real and their product equal to −1. Because y =cot x is a bijection from the interval (0,π) to the real number , there is a unique pair of numbers x_(1.1) and x_(2.1) with 0<x_(1.1) ,x_(2.1) <π such that cot x_(1.1) =q_1 and cot x_(2.1) =q_2 . Because q_1 and q_2 are negative , (π/2)<x_(1.1) ,x_(2.1) < π and so π<x_(1.1) ,x_(2.1) < 2π . Since tan x cot x=1 both tan x and cot x have period π, it follows that 1=tan x cot x=cot x cot ((π/2)−x)=cot x cot (((3π)/2)−x) 1= cot x_(1.1) . cot x_(2.1) then x_(1.1) +x_(2.1) =((3π)/2) likewise in interval (π,2π) there is unique pair number x_(1.2) and x_(2.2) satisfying the conditions of the problem with x_(1.2) +x_(2.2) =((7π)/2) Therefore x_(1.1) +x_(2.1) +x_(1.2) +x_(2.2) = ((3π)/2)+((7π)/2)=5π

$${consider}\:{the}\:{quadratic}\:{equation} \\ $$$$\mathrm{3}{q}^{\mathrm{2}} +\mathrm{8}{q}+\mathrm{3}\:=\:\mathrm{0}\:,\:{with}\:{the}\:{roots}\:{are}\:{q}_{\mathrm{1}} =\frac{−\mathrm{8}+\mathrm{2}\sqrt{\mathrm{7}}}{\mathrm{6}} \\ $$$${and}\:{q}_{\mathrm{2}} =\frac{−\mathrm{8}−\mathrm{2}\sqrt{\mathrm{7}}}{\mathrm{6}}.\:{both}\:{roots}\:{are}\:{real}\:{and} \\ $$$${their}\:{product}\:{equal}\:{to}\:−\mathrm{1}.\:{Because}\:{y}\:=\mathrm{cot}\:{x}\:{is}\:{a}\:{bijection} \\ $$$${from}\:{the}\:{interval}\:\left(\mathrm{0},\pi\right)\:{to}\:{the}\:{real}\:{number}\:,\:{there} \\ $$$${is}\:{a}\:{unique}\:{pair}\:{of}\:{numbers}\:{x}_{\mathrm{1}.\mathrm{1}} \:{and}\:{x}_{\mathrm{2}.\mathrm{1}} \\ $$$${with}\:\mathrm{0}<{x}_{\mathrm{1}.\mathrm{1}} ,{x}_{\mathrm{2}.\mathrm{1}} <\pi\:{such}\:{that}\:\mathrm{cot}\:{x}_{\mathrm{1}.\mathrm{1}} ={q}_{\mathrm{1}} \\ $$$${and}\:\mathrm{cot}\:{x}_{\mathrm{2}.\mathrm{1}} ={q}_{\mathrm{2}} .\:{Because}\:{q}_{\mathrm{1}} {and}\:{q}_{\mathrm{2}} \:{are} \\ $$$${negative}\:,\:\frac{\pi}{\mathrm{2}}<{x}_{\mathrm{1}.\mathrm{1}} \:,{x}_{\mathrm{2}.\mathrm{1}} \:<\:\pi\:{and}\:{so}\: \\ $$$$\pi<{x}_{\mathrm{1}.\mathrm{1}} \:,{x}_{\mathrm{2}.\mathrm{1}} \:<\:\mathrm{2}\pi\:.\:{Since}\:\mathrm{tan}\:{x}\:\mathrm{cot}\:{x}=\mathrm{1} \\ $$$${both}\:\mathrm{tan}\:{x}\:{and}\:\mathrm{cot}\:{x}\:{have}\:{period}\:\pi,\:{it}\:{follows}\:{that} \\ $$$$\mathrm{1}=\mathrm{tan}\:{x}\:\mathrm{cot}\:{x}=\mathrm{cot}\:{x}\:\mathrm{cot}\:\left(\frac{\pi}{\mathrm{2}}−{x}\right)=\mathrm{cot}\:{x}\:\mathrm{cot}\:\left(\frac{\mathrm{3}\pi}{\mathrm{2}}−{x}\right) \\ $$$$\mathrm{1}=\:\mathrm{cot}\:{x}_{\mathrm{1}.\mathrm{1}} .\:\mathrm{cot}\:{x}_{\mathrm{2}.\mathrm{1}} \:{then}\:{x}_{\mathrm{1}.\mathrm{1}} +{x}_{\mathrm{2}.\mathrm{1}} =\frac{\mathrm{3}\pi}{\mathrm{2}} \\ $$$${likewise}\:{in}\:{interval}\:\left(\pi,\mathrm{2}\pi\right)\:{there}\:{is}\:{unique} \\ $$$${pair}\:{number}\:{x}_{\mathrm{1}.\mathrm{2}} {and}\:{x}_{\mathrm{2}.\mathrm{2}} \:{satisfying}\:{the} \\ $$$${conditions}\:{of}\:{the}\:{problem}\:{with}\:{x}_{\mathrm{1}.\mathrm{2}} +{x}_{\mathrm{2}.\mathrm{2}} =\frac{\mathrm{7}\pi}{\mathrm{2}} \\ $$$${Therefore}\:{x}_{\mathrm{1}.\mathrm{1}} +{x}_{\mathrm{2}.\mathrm{1}} +{x}_{\mathrm{1}.\mathrm{2}} +{x}_{\mathrm{2}.\mathrm{2}} =\:\frac{\mathrm{3}\pi}{\mathrm{2}}+\frac{\mathrm{7}\pi}{\mathrm{2}}=\mathrm{5}\pi \\ $$

Answered by TANMAY PANACEA last updated on 24/Oct/20

tanx=a (3/a^2 )+(8/a)+3=0 3+8a+3a^2 =0 a=((−8±(√(8^2 −4×3×3)))/(2×3))=((−8±2(√7))/(2×3))=((−4±(√7))/3) tanx_1 =((−4+(√7))/3) =tanα tanx_2 =((−4−(√7))/3)=tanβ tan(α+β)=((tanα+tanβ)/(1−tanαtanβ))=(((−8)/3)/(1−(((16−7)/9))))=tan(π/2) (α+β)=(π/2) tan(α+β)=tan(π/2)=tan(π+(π/2))=tan((3π)/2) tanα=tan(π+α) tanβ=tan(π+β) α+(α+π)+β+(β+π)+(α+β)+(π+α+β) =4(α+β)+3π =4((π/2))+3π=5π

$${tanx}={a} \\ $$$$\frac{\mathrm{3}}{{a}^{\mathrm{2}} }+\frac{\mathrm{8}}{{a}}+\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{3}+\mathrm{8}{a}+\mathrm{3}{a}^{\mathrm{2}} =\mathrm{0} \\ $$$${a}=\frac{−\mathrm{8}\pm\sqrt{\mathrm{8}^{\mathrm{2}} −\mathrm{4}×\mathrm{3}×\mathrm{3}}}{\mathrm{2}×\mathrm{3}}=\frac{−\mathrm{8}\pm\mathrm{2}\sqrt{\mathrm{7}}}{\mathrm{2}×\mathrm{3}}=\frac{−\mathrm{4}\pm\sqrt{\mathrm{7}}}{\mathrm{3}} \\ $$$${tanx}_{\mathrm{1}} =\frac{−\mathrm{4}+\sqrt{\mathrm{7}}}{\mathrm{3}}\:={tan}\alpha\:\:\:\:\:\:\:{tanx}_{\mathrm{2}} =\frac{−\mathrm{4}−\sqrt{\mathrm{7}}}{\mathrm{3}}={tan}\beta \\ $$$${tan}\left(\alpha+\beta\right)=\frac{{tan}\alpha+{tan}\beta}{\mathrm{1}−{tan}\alpha{tan}\beta}=\frac{\frac{−\mathrm{8}}{\mathrm{3}}}{\mathrm{1}−\left(\frac{\mathrm{16}−\mathrm{7}}{\mathrm{9}}\right)}={tan}\frac{\pi}{\mathrm{2}} \\ $$$$\left(\alpha+\beta\right)=\frac{\pi}{\mathrm{2}} \\ $$$${tan}\left(\alpha+\beta\right)={tan}\frac{\pi}{\mathrm{2}}={tan}\left(\pi+\frac{\pi}{\mathrm{2}}\right)={tan}\frac{\mathrm{3}\pi}{\mathrm{2}} \\ $$$${tan}\alpha={tan}\left(\pi+\alpha\right) \\ $$$${tan}\beta={tan}\left(\pi+\beta\right) \\ $$$$\alpha+\left(\alpha+\pi\right)+\beta+\left(\beta+\pi\right)+\left(\alpha+\beta\right)+\left(\pi+\alpha+\beta\right) \\ $$$$=\mathrm{4}\left(\alpha+\beta\right)+\mathrm{3}\pi \\ $$$$=\mathrm{4}\left(\frac{\pi}{\mathrm{2}}\right)+\mathrm{3}\pi=\mathrm{5}\pi \\ $$$$ \\ $$

Commented by 1549442205PVT last updated on 24/Oct/20

Sir mistake at place (√(8^2 −4×3×3)) =(√(28))=2(√7)

$$\mathrm{Sir}\:\mathrm{mistake}\:\mathrm{at}\:\mathrm{place}\:\sqrt{\mathrm{8}^{\mathrm{2}} −\mathrm{4}×\mathrm{3}×\mathrm{3}} \\ $$$$=\sqrt{\mathrm{28}}=\mathrm{2}\sqrt{\mathrm{7}} \\ $$

Commented by TANMAY PANACEA last updated on 24/Oct/20

yes sir corrected

$${yes}\:{sir}\:{corrected} \\ $$

Answered by MJS_new last updated on 24/Oct/20

3cot^2 x +8cot x +3=0 ((3cos^2 x)/(sin^2 x))+((8cos x)/(sin x))+3=0 ((3cos^2 +8cos x sin x +3sin^2 x)/(sin^2 x))=0 3+8cos x sin x =0 cos x sin x =−(3/8) (1/2)sin 2x =−(3/8) sin 2x =−(3/4) x=nπ−(1/2)arcsin (3/4) ∨ x=nπ+(π/2)+(1/2)arcsin (3/4) let α=(1/2)arcsin (3/4) for 0≤x<2π we get x=π−α∨x=2π−α∨x=(π/2)+α∨x=((3π)/2)+α ⇒ sum of these is 5π

$$\mathrm{3cot}^{\mathrm{2}} \:{x}\:+\mathrm{8cot}\:{x}\:+\mathrm{3}=\mathrm{0} \\ $$$$\frac{\mathrm{3cos}^{\mathrm{2}} \:{x}}{\mathrm{sin}^{\mathrm{2}} \:{x}}+\frac{\mathrm{8cos}\:{x}}{\mathrm{sin}\:{x}}+\mathrm{3}=\mathrm{0} \\ $$$$\frac{\mathrm{3cos}^{\mathrm{2}} +\mathrm{8cos}\:{x}\:\mathrm{sin}\:{x}\:+\mathrm{3sin}^{\mathrm{2}} \:{x}}{\mathrm{sin}^{\mathrm{2}} \:{x}}=\mathrm{0} \\ $$$$\mathrm{3}+\mathrm{8cos}\:{x}\:\mathrm{sin}\:{x}\:=\mathrm{0} \\ $$$$\mathrm{cos}\:{x}\:\mathrm{sin}\:{x}\:=−\frac{\mathrm{3}}{\mathrm{8}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{2}{x}\:=−\frac{\mathrm{3}}{\mathrm{8}} \\ $$$$\mathrm{sin}\:\mathrm{2}{x}\:=−\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${x}={n}\pi−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arcsin}\:\frac{\mathrm{3}}{\mathrm{4}}\:\vee\:{x}={n}\pi+\frac{\pi}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arcsin}\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\mathrm{let}\:\alpha=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arcsin}\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\mathrm{for}\:\mathrm{0}\leqslant{x}<\mathrm{2}\pi\:\mathrm{we}\:\mathrm{get} \\ $$$${x}=\pi−\alpha\vee{x}=\mathrm{2}\pi−\alpha\vee{x}=\frac{\pi}{\mathrm{2}}+\alpha\vee{x}=\frac{\mathrm{3}\pi}{\mathrm{2}}+\alpha \\ $$$$\Rightarrow\:\mathrm{sum}\:\mathrm{of}\:\mathrm{these}\:\mathrm{is}\:\mathrm{5}\pi \\ $$

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°

$$\mathrm{3cot}\:^{\mathrm{2}} {x}\:+\:\mathrm{8}\:\mathrm{cot}\:{x}\:+\:\mathrm{3}\:=\:\mathrm{0} \\ $$$$\Delta'=\mathrm{4}^{\mathrm{2}} −\mathrm{9}=\mathrm{7}\Rightarrow\mathrm{cotx}=\frac{−\mathrm{4}\pm\sqrt{\mathrm{7}}}{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{tanx}=\frac{\mathrm{3}}{−\mathrm{4}\pm\sqrt{\mathrm{7}}}=\frac{\mathrm{3}\left(−\mathrm{4}\mp\sqrt{\mathrm{7}}\right)}{\mathrm{9}}=\frac{−\mathrm{4}\mp\sqrt{\mathrm{7}}}{\mathrm{3}} \\ $$$$\mathrm{Put}\:\alpha=\mathrm{tan}^{−\mathrm{1}} \left(\frac{−\mathrm{4}+\sqrt{\mathrm{7}}}{\mathrm{3}}\right),\beta=\mathrm{tan}^{−\mathrm{1}} \left(\frac{−\mathrm{4}−\sqrt{\mathrm{7}}}{\mathrm{3}}\right) \\ $$$$\alpha\approx−\mathrm{24}°\mathrm{17}'\mathrm{42}'',\beta\approx−\mathrm{65}°\mathrm{42}'\mathrm{17}'' \\ $$$$\mathrm{Since}\:\alpha,\beta\in\left[\mathrm{0},\mathrm{2}\pi\right]\:\mathrm{and}\:\mathrm{tanx}=\mathrm{tan}\left(\mathrm{x}+\mathrm{k}\pi\right) \\ $$$$\mathrm{we}\:\mathrm{get}\:\alpha\in\left\{\mathrm{155}°\mathrm{42}'\mathrm{17}'',\mathrm{335}°\mathrm{42}'\mathrm{17}''\right\} \\ $$$$\beta\in\left\{\mathrm{114}°\mathrm{17}'\mathrm{43}'',\mathrm{294}°\mathrm{17}'\mathrm{43}''\right\} \\ $$$$\Rightarrow\alpha_{\mathrm{1}} +\alpha_{\mathrm{2}} +\beta_{\mathrm{1}} +\beta_{\mathrm{2}} =\mathrm{900}° \\ $$$$\mathrm{Thus},\mathrm{Sum}\:\mathrm{of}\:\mathrm{all}\:\mathrm{the}\:\mathrm{angles}\:\mathrm{satisfying} \\ $$$$\mathrm{given}\:\mathrm{equation}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{900}° \\ $$

Answered by Bird last updated on 24/Oct/20

3cotan^2 x+8cotanx +3=0 ⇒ (3/(tan^2 x)) +(8/(tanx)) +3 =0 ⇒ 3+8tanx +3tan^2 x =0 ⇒ 3tan^2 x+8tanx +3 =0 tanx=u ⇒3u^2 +8u+3=0 Δ^′ =4^2 −9 =16−9 =7 ⇒ u_1 =((−4+(√7))/3) and u_2 =((−4−(√7))/3) tanx=((−4+(√7))/3) ⇒x=arctan(((−4+(√7))/3))+nπ tanx=((−4−(√7))/3) ⇒x=−arctan(((4+(√7))/3))+nπ (n∈Z)

$$\mathrm{3}{cotan}^{\mathrm{2}} {x}+\mathrm{8}{cotanx}\:+\mathrm{3}=\mathrm{0}\:\Rightarrow \\ $$$$\frac{\mathrm{3}}{{tan}^{\mathrm{2}} {x}}\:+\frac{\mathrm{8}}{{tanx}}\:+\mathrm{3}\:=\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{3}+\mathrm{8}{tanx}\:+\mathrm{3}{tan}^{\mathrm{2}} {x}\:=\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{3}{tan}^{\mathrm{2}} {x}+\mathrm{8}{tanx}\:+\mathrm{3}\:=\mathrm{0} \\ $$$${tanx}={u}\:\Rightarrow\mathrm{3}{u}^{\mathrm{2}} +\mathrm{8}{u}+\mathrm{3}=\mathrm{0} \\ $$$$\Delta^{'} \:=\mathrm{4}^{\mathrm{2}} −\mathrm{9}\:=\mathrm{16}−\mathrm{9}\:=\mathrm{7}\:\Rightarrow \\ $$$${u}_{\mathrm{1}} =\frac{−\mathrm{4}+\sqrt{\mathrm{7}}}{\mathrm{3}}\:{and}\:{u}_{\mathrm{2}} =\frac{−\mathrm{4}−\sqrt{\mathrm{7}}}{\mathrm{3}} \\ $$$${tanx}=\frac{−\mathrm{4}+\sqrt{\mathrm{7}}}{\mathrm{3}}\:\Rightarrow{x}={arctan}\left(\frac{−\mathrm{4}+\sqrt{\mathrm{7}}}{\mathrm{3}}\right)+{n}\pi \\ $$$${tanx}=\frac{−\mathrm{4}−\sqrt{\mathrm{7}}}{\mathrm{3}}\:\Rightarrow{x}=−{arctan}\left(\frac{\mathrm{4}+\sqrt{\mathrm{7}}}{\mathrm{3}}\right)+{n}\pi \\ $$$$\left({n}\in{Z}\right) \\ $$

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