find General solution cot x+cot 2x+cot3x= 0


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Question Number 105619 by bobhans last updated on 30/Jul/20

$f i n d G e n e r a l s o l u t i o n \cot x + \cot 2 x + \cot 3 x = 0$
	Answered by bemath last updated on 30/Jul/20
	$⇔cot (x)+((cot^2 (x)−1)/(2cot (x)))+((3cot (x)−cot^3 (x))/(1−3cot^2 (x)))= 0 set cot (x)= p p + ((p^2 −1)/(2p))+((3p−p^3 )/(1−3p^2 )) = 0 11(p^2 )^2 −12p^2 +1 = 0 p^2 = ((12±(√(12^2 −4×11)))/(22)) = ((12±10)/(22)) → { ((p^2 =1 ; p = ±1)),((p^2 = (1/(11)); p = ±(1/(√(11))))) :} → { ((tan (x)= ±1)),((tan (x)= ±(√(11)))) :} → { ((x=±(π/4)+k.π)),((x= ±arc tan ((√(11)))+k.π)) :}$
	$\Leftrightarrow \cot (x) + \frac{\cot^{2} (x) - 1}{2 \cot (x)} + \frac{3 \cot (x) - \cot^{3} (x)}{1 - 3 \cot^{2} (x)} = 0$ $s e t \cot (x) = p$ $p + \frac{p^{2} - 1}{2 p} + \frac{3 p - p^{3}}{1 - 3 p^{2}} = 0$ $11 {(p^{2})}^{2} - 12 p^{2} + 1 = 0$ $p^{2} = \frac{12 \pm \sqrt{12^{2} - 4 \times 11}}{22} = \frac{12 \pm 10}{22}$ $\to {\begin{cases} p^{2} = 1; p = \pm 1 \\ p^{2} = \frac{1}{11}; p = \pm \frac{1}{\sqrt{11}} \end{cases}$ $\to {\begin{cases} \tan (x) = \pm 1 \\ \tan (x) = \pm \sqrt{11} \end{cases}$ $\to {\begin{cases} x = \pm \frac{π}{4} + k . π \\ x = \pm arc \tan (\sqrt{11}) + k . π \end{cases}$
	Answered by 1549442205PVT last updated on 30/Jul/20
	$cot x+cot 2x+cot3x= 0 (1) we need the conditions: { ((sinx≠0)),((sin2x≠0)),((sin3x≠0)) :} ⇔x≠kπ,x≠((mπ)/2),x≠((nπ)/3)⇔ { ((x≠((nπ)/3))),((x≠(π/2)+kπ)) :} (1)⇔((cosx)/(sinx))+((cos2x)/(sim2x))+((cos3x)/(sin3x))=0 ⇔((sin3xcosx+cos3xsinx)/(sinxsin3x))+((cos2x)/(sin2x))=0 ⇔((sin4x)/(sinxsin3x))+((cos2x)/(sin2x))=0⇔sin4xsin2x+cos2xsinxsin3x=0 2sin^2 2xcos2x+cos2xsinxsin3x=0 ⇔cos2x(2sin^2 2x+sinxsin3x)=0 i)cos2x=0⇔2x=k(π/2)⇔x=((kπ)/4) ii)2sin^2 2x+sinxsin3x=0⇔8sin^2 xcos^2 x+sinx(3sinx−4sin^3 x)=0 ⇔sin^2 x(8cos^2 x+3−4sin^2 x)=0 ⇔8cos^2 x+3−4(1−cos^2 x)=0 (as sinx≠0) ⇔12cos^2 x−1=0⇔cos^2 x=(1/(12))⇔cosx=±((√3)/6) ⇔x=±cos^(−1) (((√3)/6))+2mπ or x=±cos^(−1) (−((√3)/6)) Thus,the solutions of the given equation are: x∈{((k𝛑)/4);±cos^(−1) (((√3)/6))+2m𝛑;±cos^(−1) (((−(√3))/6))+2n𝛑} where k≠4p,k≠4q+2(p,q∈Z)$
	$\cot x + \cot 2 x + \cot 3 x = 0 (1)$ $we need the conditions : {\begin{cases} sinx \neq 0 \\ \sin 2 x \neq 0 \\ \sin 3 x \neq 0 \end{cases}$ $\Leftrightarrow x \neq k π, x \neq \frac{m π}{2}, x \neq \frac{n π}{3} \Leftrightarrow {\begin{cases} x \neq \frac{n π}{3} \\ x \neq \frac{π}{2} + k π \end{cases}$ $(1) \Leftrightarrow \frac{cosx}{sinx} + \frac{\cos 2 x}{sim 2 x} + \frac{\cos 3 x}{\sin 3 x} = 0$ $\Leftrightarrow \frac{\sin 3 xcosx + \cos 3 xsinx}{sinxsin 3 x} + \frac{\cos 2 x}{\sin 2 x} = 0$ $\Leftrightarrow \frac{\sin 4 x}{sinxsin 3 x} + \frac{\cos 2 x}{\sin 2 x} = 0 \Leftrightarrow \sin 4 xsin 2 x + \cos 2 xsinxsin 3 x = 0$ ${2 \sin}^{2} 2 xcos 2 x + \cos 2 xsinxsin 3 x = 0$ $\Leftrightarrow \cos 2 x ({2 \sin}^{2} 2 x + sinxsin 3 x) = 0$ $i) \cos 2 x = 0 \Leftrightarrow 2 x = k \frac{π}{2} \Leftrightarrow x = \frac{k π}{4}$ $ii) {2 \sin}^{2} 2 x + sinxsin 3 x = 0 \Leftrightarrow {8 \sin}^{2} {xcos}^{2} x + sinx (3 sinx - {4 \sin}^{3} x) = 0$ $\Leftrightarrow \sin^{2} x ({8 \cos}^{2} x + 3 - {4 \sin}^{2} x) = 0$ $\Leftrightarrow {8 \cos}^{2} x + 3 - 4 (1 - \cos^{2} x) = 0 (as sinx \neq 0)$ $\Leftrightarrow {12 \cos}^{2} x - 1 = 0 \Leftrightarrow \cos^{2} x = \frac{1}{12} \Leftrightarrow cosx = \pm \frac{\sqrt{3}}{6}$ $\Leftrightarrow x = \pm \cos^{- 1} (\frac{\sqrt{3}}{6}) + 2 m π or x = \pm \cos^{- 1} (- \frac{\sqrt{3}}{6})$ $Thus, the solutions of the given equation are :$ $x \in {\frac{k π}{4}; \pm \cos^{- 1} (\frac{\sqrt{3}}{6}) + 2 m π; \pm \cos^{- 1} (\frac{- \sqrt{3}}{6}) + 2 n π}$ $where k \neq 4 p, k \neq 4 q + 2 (p, q \in Z)$
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