Solve-for-x-tan-x-tan-2x-tan-3x-0-Some-one-had-posted-this-question-and-I-had-answered-it-but-then-thread-was-deleted-I-think-that-the-question-is-not-importanceless-so-I-hav-reposted-it-

FacebookTweetPin

Question Number 1890 by Rasheed Soomro last updated on 21/Oct/15

$$Solveforx$$$$tanx+tan2x-tan3x=0$$$$SomeonehadpostedthisquestionandIhadansweredit$$$$butthenthreadwasdeleted!$$$$Ithinkthatthequestionisnotimportanceless,soIhav$$$$repostedit.$$

Commented by Rasheed Soomro last updated on 22/Oct/15

$$Question{sholdn}^{\prime }tbedeleted/changedincaseitisanswered/commented.$$$$Becauseafterbeinganswered/commenteditisnolonger$$$$remainedyourprivatematter.$$

Commented by 123456 last updated on 21/Oct/15

$$\mathrm{all}\mathrm{questions}\mathrm{are}\mathrm{importants}:3$$

Commented by Rasheed Soomro last updated on 23/Oct/15

$$ofcourse!$$

Answered by Yozzy last updated on 22/Oct/15

$$Onemethodofapproachinvolvestheuseofthefollowingtrigonometricidentity:$$$$tan3x=\frac{tan2x+tanx}{1-tan2xtanx}\dots \dots \dots \dots ..\left(1\right)$$$$Thegivenequationthentakestheform,afterfactorising,$$$$(tanx+tan2x)(1-\frac{1}{1-tanxtan2x})=0\{tanxtan2x\ne 1\}$$$$(tanx+tan2x)\left(\frac{1-tanxtan2x-1}{1-tanxtan2x}\right)=0$$$$\frac{tanxtan2x(tanx+tan2x)}{tanxtan2x-1}=0\dots \dots \dots \dots \dots ..\left(2\right)$$$$For\left(2\right)tobetruewehavethepossibilityoftanx=0ortan2x=0ortanx+tan2x=0.$$$$Iftanx=0wehaveageneralsolutionasx=n\pi ,n\in \mathbb{Z}.$$$$Iftan2x=0wehaveageneralsolutionasx=\frac{n\pi }{2},n\in \mathbb{Z}.$$$$Iftanx+tan2x=0\Rightarrow tan2x=-tanx\Rightarrow 2x=-x+n\pi \Rightarrow x=\frac{n\pi }{3},n\in \mathbb{Z}.$$$$Tocheckthelastsolutionfunction\left(sincexisthenafunctionofn\right)observe$$$$thatfrom\left(1\right)$$$$tanx+tan2x=tan3x(1-tanxtan2x).$$$$Havingagreeduponthattanxtan2x\ne 1\Rightarrow tanxtan2x-1\ne 0.Thereforeweought$$$$tohavetan3x=0\Rightarrow 3x=n\pi \Rightarrow x=\frac{n\pi }{3},n\in \mathbb{Z}.$$$$Hence,apossiblegeneralsolutionisgivenas$$$$S=\{x\in \mathbb{R},n,m,q\in \mathbb{Z}\mid x=n\pi \vee x=\frac{m\pi }{2}\vee x=\frac{q\pi }{3}\}.$$$$However,ifmisoddwethattheequationisnotsatifiedsincetan\frac{m\pi }{2}isundefined$$$$formoddsothatwehavethetruesolutionset$$$$S=\{x\in \mathbb{R},n,q\in \mathbb{Z}\mid x=n\pi \vee x=\frac{q\pi }{3}\}.$$$$Iftanxtan2x=1\Rightarrow 2{tan}^{2}x=1-{tan}^{2}x\Rightarrow {tan}^{2}x=\frac{1}{3}\Rightarrow tanx=\pm \frac{1}{\sqrt{3}}\Rightarrow x=\{\begin{array}{l}\left(\frac{6n+1}{6}\right)\pi n\in \mathbb{Z}\\ \left(\frac{6n-1}{6}\right)\pi n\in \mathbb{Z}\end{array}$$$$Theabovegeneralsolutionincursasolutionset{S}^{{}^{\prime }}completelydisjointfrom$$$$Ssince6n\pm 1isodd\mathrm{\forall }n\in \mathbb{Z}\Rightarrow 6neverdivides6n\pm 1toyieldaquotientthatisan$$$$integralmultipleof1,\frac{1}{2}and\frac{1}{3}.Therefore,Sappearstobethecompletesetofreal$$$$solutions.Iwonderwhatcomplexsolutionsexistfortheequation.$$$$$$$$$$$$$$$$$$

Answered by Rasheed Soomro last updated on 22/Oct/15

$$EasyandShortapproach$$$$tanx+tan2x-tan3x=0$$$$tanx+tan2x-tan(2x+x)=0$$$$tanx+tan2x-\frac{tan2x+tanx}{1-tan2xtanx}=0$$$$Multiplyingbothsidesby1-tan2xtanx:$$$$\stackrel{\times }{tanx}-tan2x{tan}^{2}x+\stackrel{\times }{tan2x}-{tan}^{2}2xtanx-\stackrel{\times }{tan2x}-\stackrel{\times }{tanx}=0$$$$-tan2x{tan}^{2}x-{tan}^{2}2xtanx=0$$$$tan2x{tan}^{2}x+{tan}^{2}2xtanx=0$$$$tan2xtanx(tanx+tan2x)=0$$$$tan2x=0\mid tanx=0\mid tanx+tan2x=0$$$$tanx=0\Rightarrow x=n\pi$$$$tan2x=0\Rightarrow x=\frac{n\pi }{2}$$$$tanx+tan2x=0\Rightarrow tanx=-tan2x\Rightarrow tanx=tan(-2x)$$$$\Rightarrow x=-2x+n\pi \Rightarrow x=\frac{n\pi }{3}$$$$SolutionSetforx$$$$\{n\pi ,\frac{n\pi }{2},\frac{n\pi }{3}\}$$$$Note:Insolvingtanx+tan2x=0Ihavegothelpfrom$$$$theaboveanswerbyYozzy.$$

Terms of Service
Privacy Policy
Contact: info@tinkutara.com

FacebookTweetPin