How do you solve ${tan}^{- 1} (2 x) + {tan}^{- 1} (x) = \frac{3 π}{17}$ $tan^-1(2x)+tan^-1(x)= (3pi)/17$ ?

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How do you solve ${tan}^{- 1} (2 x) + {tan}^{- 1} (x) = \frac{3 π}{17}$ $tan^-1(2x)+tan^-1(x)= (3pi)/17$ ?

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Answer 1 · 2016-05-25T04:33:53

$x = \frac{π}{17}$ $x = pi/17$

Explanation:

${tan}^{- 1} (2 x)$ $tan^-1 (2x)$ --> arc (2x)
${tan}^{- 1} x$ $tan ^-1 x$ --> arc x
The equation is transformed to:
$2 x + x = 3 x = \frac{3 π}{17}$ $2x + x = 3x = (3pi)/17$
$x = \frac{π}{17}$ $x = pi/17$
Answers in degrees:
$\frac{π}{17} = 10^{\circ} 59$ $pi/17 = 10^@59$
$\frac{2 π}{17} = 21^{\circ} 18$ $(2pi)/17 = 21^@18$

Answer 2 · 2016-05-25T16:38:05

Let $c = tan (\frac{3 π}{17}) = 0.619174$ $c = tan ((3pi)/17) = 0.619174$ , nearly. Then, $x = \frac{- 3 \pm \sqrt{9 + 8 c^{2}}}{4 c} = 0.19129 and - 2.61367$ $x =(-3+- sqrt(9+8c^2))/(4c)=0.19129 and -2.61367$ , nearly It is seemingly Incredible that x could be negative.

Explanation:

Let $a = {tan}^{- 1} (2 x) and b = {tan}^{- 1} (x)$ $a = tan^(-1)(2x) and b = tan^(-1) (x)$ .

Then $tan (a + b) = \frac{tan a + tan b}{1 - t n a tan b}$ $tan (a + b ) = (tan a + tan b )/(1-tna tan b )$

$= \frac{2 x + x}{1 - (2 x) (x)}$ $=(2x+x)/(1-(2x)(x))$

$= \frac{3 x}{1 - 2 x^{2}} = tan (\frac{3 π}{17}) = c = 0.619174$ $=(3x)/(1-2x^2) = tan ((3pi)/17) = c=0.619174$ , nearly.

Solving this quadratic for a positive x,

$x = \frac{\sqrt{9 + 8 c^{2}} - 3}{4 c} = 0.19129$ $x = (sqrt(9+8c^2)-3)/(4c)=0.19129$ , nearly.

Here, the two angles that add up to $\frac{3 π}{17} = {31.76}^{o}$ $(3pi)/17 =31.76^o$ are #20.94^o

and 10.83^o#.

Seemingly incredible but true, the negative root $= - 2.61367$ $=-2.61367$ is

also a solution.

The corresponding angles are ${100.83}^{o} and - {69.06}^{o}$ $100.83^o and -69.06^o$ .

I have used that tangent is negative in the 2nd quadrant for

${tan}^{- 1} (2 x)$ $tan^(-1)(2x)$ and in the fourth quadrant for ${tan}^{- 1} (x)$ $tan^(-1)(x)$ , respectively.

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How do you solve ${tan}^{- 1} (2 x) + {tan}^{- 1} (x) = \frac{3 π}{17}$ $tan^-1(2x)+tan^-1(x)= (3pi)/17$ ?

2 Answers

Explanation:

Explanation:

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How do you solve tan^-1(2x)+tan^-1(x)= (3pi)/17tan−1(2x)+tan−1(x)=3π17tan^-1(2x)+tan^-1(x)= (3pi)/17?

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Explanation:

Explanation:

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How do you solve ${tan}^{- 1} (2 x) + {tan}^{- 1} (x) = \frac{3 π}{17}$ $tan^-1(2x)+tan^-1(x)= (3pi)/17$ ?