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

2 Answers
Nghi N.
May 25, 2016

x = pi/17

Explanation:

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

A. S. Adikesavan · sente
May 25, 2016

Let c = tan ((3pi)/17) = 0.619174, nearly. Then, 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)(2x) and b = tan^(-1) (x).

Then tan (a + b ) = (tan a + tan b )/(1-tna tan b )

=(2x+x)/(1-(2x)(x))

=(3x)/(1-2x^2) = tan ((3pi)/17) = c=0.619174, nearly.

Solving this quadratic for a positive x,

x = (sqrt(9+8c^2)-3)/(4c)=0.19129, nearly.

Here, the two angles that add up to (3pi)/17 =31.76^o are #20.94^o

and 10.83^o#.

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

also a solution.

The corresponding angles are 100.83^o and -69.06^o.

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

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

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