Question

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

Answer 1

`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`

Answer 2

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.