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

Question Number 109818 by Study last updated on 25/Aug/20

t a n x + t a n 2 x = 1

f i n d t h e g e n e r a l s o l u t i o n

Commented by Dwaipayan Shikari last updated on 25/Aug/20

x = k π + \frac{261}{720} π (a p p r o x i m a t e l y) (k \in Z)

Commented by Study last updated on 25/Aug/20

w h a t i s t h e p r a c t i c e ?

Commented by Dwaipayan Shikari last updated on 25/Aug/20

t a n x + \frac{2 t a n x}{1 - {t a n}^{2} x} = 1

t + \frac{2 t}{1 - t^{2}} = 1

t - t^{3} + 2 t = 1 - t^{2}

t^{3} - t^{2} - 3 t + 1 = 0

t = 2.17 \dots .

t = a n o t h e r s o l u t i o n

t a n x = 2.17 \dots

x = k π + \frac{261}{720} π (a p p r o x)

Commented by Study last updated on 25/Aug/20

h e l p m e

Commented by Study last updated on 25/Aug/20

i n e e d y o u r h e l p

Commented by Study last updated on 25/Aug/20

h o w w e c a n f i n d t h e t v a l u e f r o m t h e

t h i r d d e g r e e e q u a t i o n ?

Commented by Her_Majesty last updated on 25/Aug/20

i n t h i s c a s e y o u n e e d t h e t r i g o n o m e t r i c

s o l u t i o n m e t h o d . s e a r c h f o r i t o n t h e w e b

a n y w a y b e t t e r u s e a p p r o x i m a t i o n

w e g e t

t_{1} \approx - 1.48119

t_{2} \approx . 311108

t_{3} \approx 2.17001

\Rightarrow

x_{1} = - . 976957 + n π

x_{2} = . 301616 + n π

x_{3} = 1.13899 + n π

Answered by 1549442205PVT last updated on 26/Aug/20

We need the condition x \neq \frac{π}{2}, x \neq \frac{π}{4}

tanx + \frac{2 tanx}{1 - \tan^{2} x} = 1 \Leftrightarrow tanx - \tan^{3} x + 2 tanx

= 1 - \tan^{2} x \Leftrightarrow \tan^{3} x - \tan^{2} x - 3 tanx + 1 = 0

\Rightarrow \cot^{3} x - {3 \cot}^{2} x - cotx + 1 = 0

{(cotx - 1)}^{3} - 4 (cotx - 1) - 2 = 0

Put cotx - 1 = y we get

y^{3} - 4 y - 2 = 0 (2)

Set y = \frac{4}{\sqrt{3}} z . Replace into (2) we get

\frac{64}{3 \sqrt{3}} z^{3} - \frac{16}{\sqrt{3}} z - 2 = 0 . Multiplying two

sides by \frac{3 \sqrt{3}}{16} we obtain :

{4 z}^{3} - 3 z - \frac{3 \sqrt{3}}{8} = 0 (*) . On the other hands,

we known that {4 \cos}^{3} u - 3 cosu - \cos 3 u = 0

Hence, Put \cos 3 u = \frac{3 \sqrt{3}}{8} (3) then we see

z = cosu is roots of the equation (*)

(3) \Leftrightarrow 3 u = \cos^{- 1} (\frac{3 \sqrt{3}}{8}) + 2 k π \Leftrightarrow u = \frac{1}{3} [\cos^{- 1} (\frac{3 \sqrt{3}}{8}) + 2 k π]

Thus, the equation (*) has three roots

are z = \cos \frac{1}{3} [\cos^{- 1} (\frac{3 \sqrt{3}}{8}) + 2 k π] (k = 0, 1, 2)

\Rightarrow y = \frac{4}{\sqrt{3}} z = \frac{4}{\sqrt{3}} \cos \frac{1}{3} [\cos^{- 1} (\frac{3 \sqrt{3}}{8}) + 2 k π]

\Rightarrow cotx = y + 1 = 1 + \frac{4}{\sqrt{3}} \cos \frac{1}{3} [\cos^{- 1} (\frac{3 \sqrt{3}}{8}) + 2 k π]

\Rightarrow tanx = \frac{1}{1 + \frac{4}{\sqrt{3}} \cos \frac{1}{3} [\cos^{- 1} (\frac{3 \sqrt{3}}{8}) + 2 k π]}

Thus, the given equation has roots are

x = \tan^{- 1} {\frac{1}{1 + \frac{4}{\sqrt{3}} \cos \frac{1}{3} [\cos^{- 1} (\frac{3 \sqrt{3}}{8}) + 2 k π]}} + m π

where k = 0, 1, 2 and m \in Z