find-General-solution-cot-x-cot-2x-cot3x-0-

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

\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 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)