ARML-Inspired Equation

Our main solution here is to use the trigonometric identity (this is yours to prove since this is trivial by using the tan addition identity):

arctan (a) + arctan (b) = arctan([a+b]/[1 - ab]).

We can separate the terms in order to use the identity given above, where

4 arctan (x) + 4 arctan (3x) + 2 arctan (3x) = pi

4 (arctan [4x / (1 - 3x^2)]) + 2 arctan (3x) = pi

Separating further the terms:

2 (arctan [4x / (1 - 3x^2)]) + 2 (arctan [4x / (1 - 3x^2)]) + 2 arctan (3x) = pi

2 (arctan [4x / (1 - 3x^2)]) + 2 arctan ([7x - 9x^3]/[1 - 15x^2]) = pi

2 arctan ([27x^5 - 90x^3 + 11x] / [81x^4 - 46x^2 + 1]) = pi

Note: The operations here are all simplified. It is your work to use these identities and simplify after adding and dividing.

arctan ([27x^5 - 90x^3 + 11x] / [81x^4 - 46x^2 + 1]) = pi/2

[27x^5 - 90x^3 + 11x] / [81x^4 - 46x^2 + 1] = tan(pi/2) = undefined

There exists such x^2 that will make the equation undefined and this is when 81x^4 - 46x^2 + 1 = 0 where for all x's is a vertical asymptote and noticing that the common factor between the numerator and denominator is only 1. Hence, we solve the quadratic equation above for x^2.

We call the quadratic formula to solve for x^2.

x^2 = [23 +- 8sqrt(7)]/81 after simplification.

Hence, 119 is the answer.

$Let\quad \tan ^{ -1 }{ x=\theta \quad \& } \quad \tan ^{ -1 }{ 3x=\alpha } ,Given\\ \\ 2\theta +3\alpha =\cfrac { \pi }{ 2 } \\ \\ \Rightarrow \tan { 2\theta } \tan { 3\alpha = } 1\\ \\ \Rightarrow \cfrac { 2x }{ 1-{ x }^{ 2 } } \times \cfrac { 3(3x)-{ (3x) }^{ 3 } }{ 1-3{ (3x) }^{ 2 } } =1\\ \\ \Rightarrow 81{ x }^{ 4 }-46{ x }^{ 2 }+1=0\\ \\ \Rightarrow { x }^{ 2 }=\cfrac { 46\pm \sqrt { { 46 }^{ 2 }-4\times 81\times 1 } }{ 2\times 81 } =\cfrac { 23\pm 8\sqrt { 7 } }{ 81 } =\cfrac { a-b\sqrt { c } }{ d } \\ \\ As\quad all\quad are\quad +ve\\ \\ a=23,b=8,c=7,d=81\Rightarrow a+b+c+d=119\\ \\$