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tanx-tan2x-1-find-the-general-solution-




Question Number 109818 by Study last updated on 25/Aug/20
tanx+tan2x=1  find the general solution
tanx+tan2x=1findthegeneralsolution
Commented by Dwaipayan Shikari last updated on 25/Aug/20
x=kπ+((261)/(720))π  (approximately)  (k∈Z)
x=kπ+261720π(approximately)(kZ)
Commented by Study last updated on 25/Aug/20
what is the practice?
whatisthepractice?
Commented by Dwaipayan Shikari last updated on 25/Aug/20
tanx+((2tanx)/(1−tan^2 x))=1  t+((2t)/(1−t^2 ))=1  t−t^3 +2t=1−t^2   t^3 −t^2 −3t+1=0  t=2.17....  t=another solution  tanx=2.17...  x=kπ+((261)/(720))π   (approx)
tanx+2tanx1tan2x=1t+2t1t2=1tt3+2t=1t2t3t23t+1=0t=2.17.t=anothersolutiontanx=2.17x=kπ+261720π(approx)
Commented by Study last updated on 25/Aug/20
help me
helpme
Commented by Study last updated on 25/Aug/20
i need your help
ineedyourhelp
Commented by Study last updated on 25/Aug/20
how we can find the t value from the  third degree equation?
howwecanfindthetvaluefromthethirddegreeequation?
Commented by Her_Majesty last updated on 25/Aug/20
in this case you need the trigonometric  solution method. search for it on the web  anyway better use approximation  we get  t_1 ≈−1.48119  t_2 ≈.311108  t_3 ≈2.17001  ⇒  x_1 =−.976957+nπ  x_2 =.301616+nπ  x_3 =1.13899+nπ
inthiscaseyouneedthetrigonometricsolutionmethod.searchforitonthewebanywaybetteruseapproximationwegett11.48119t2.311108t32.17001x1=.976957+nπx2=.301616+nπx3=1.13899+nπ
Answered by 1549442205PVT last updated on 26/Aug/20
We need the condition x≠(π/2),x≠(π/4)  tanx+((2tanx)/(1−tan^2 x))=1⇔tanx−tan^3 x+2tanx  =1−tan^2 x⇔tan^3 x−tan^2 x−3tanx+1=0  ⇒cot^3 x−3cot^2 x−cotx+1=0  (cotx−1)^3 −4(cotx−1)−2=0  Put cotx−1=y we get  y^3 −4y−2=0(2)  Set y=(4/( (√3)))z.Replace into (2)we get  ((64)/(3(√3)))z^3 −((16)/( (√3)))z−2=0.Multiplying two  sides by ((3(√3))/(16)) we obtain:  4z^3 −3z−((3(√3))/8)=0(∗).On the other hands,  we known that 4cos^3 u−3cosu−cos3u=0  Hence ,Put cos3u=((3(√3))/8)(3) then we see  z=cosu is roots of the equation (∗)  (3)⇔3u=cos^(−1) (((3(√3))/8))+2kπ⇔u=(1/3)[cos^(−1) (((3(√3))/8))+2kπ]  Thus,the equation (∗)has three roots  are z=cos(1/3)[cos^(−1) (((3(√3))/8))+2kπ](k=0,1,2)  ⇒y=(4/( (√3)))z=(4/( (√3)))cos(1/3)[cos^(−1) (((3(√3))/8))+2kπ]  ⇒cotx=y+1=1+(4/( (√3)))cos(1/3)[cos^(−1) (((3(√3))/8))+2kπ]  ⇒tanx=(1/(1+(4/( (√3)))cos(1/3)[cos^(−1) (((3(√3))/8))+2kπ]))  Thus,the given equation has roots are  x=tan^(−1) {(1/(1+(4/( (√3)))cos(1/3)[cos^(−1) (((3(√3))/8))+2kπ]))}+mπ  where k=0,1,2 and m∈Z
Weneedtheconditionxπ2,xπ4tanx+2tanx1tan2x=1tanxtan3x+2tanx=1tan2xtan3xtan2x3tanx+1=0cot3x3cot2xcotx+1=0(cotx1)34(cotx1)2=0Putcotx1=ywegety34y2=0(2)Sety=43z.Replaceinto(2)weget6433z3163z2=0.Multiplyingtwosidesby3316weobtain:4z33z338=0().Ontheotherhands,weknownthat4cos3u3cosucos3u=0Hence,Putcos3u=338(3)thenweseez=cosuisrootsoftheequation()(3)3u=cos1(338)+2kπu=13[cos1(338)+2kπ]Thus,theequation()hasthreerootsarez=cos13[cos1(338)+2kπ](k=0,1,2)y=43z=43cos13[cos1(338)+2kπ]cotx=y+1=1+43cos13[cos1(338)+2kπ]tanx=11+43cos13[cos1(338)+2kπ]Thus,thegivenequationhasrootsarex=tan1{11+43cos13[cos1(338)+2kπ]}+mπwherek=0,1,2andmZ

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