What is x if tan^-1(2x)+tan^-1(3x)=pi/4?

We'll write tan^-1(x) = arctan
x

We'll re-write the
equation:

arctan 2x + arctan 3x = `pi`
/4

We'll take tangent function both
sides:

tan(arctan 2x + arctan 3x) = tan `pi`
/4

We'll use the
formula:

tan(a+b) = (tan a + tan b)/(1 - tan a*tan
b)

tan(arctan 2x + arctan 3x) = (tan (arctan 2x) + tan
(arctan 3x))/(1 - tan (arctan 2x)*tan (arctan 3x))

But
tan(arctanx) = x

tan(arctan 2x + arctan 3x) =
(2x+3x)/(1-6x^2)

We'll re-write the
equation:

(2x+3x)/(1-6x^2) =
1

We'll subtract 1 both
sides:

(2x+3x)/(1-6x^2) - 1 =
0

(2x + 3x - 1 + 6x^2)/(1-6x^2) =
0

Since the denominator must not be zero, then only the
numerator can cancel the fraction.

6x^2 + 5x - 1 =
0

We'll apply quadratic
formula:

x1 = [-5+sqrt(25 +
24)]/12

x1 = (-5 + 7)/12

x1 =
2/12

x1 = 1/6

x2 = (-5 -
7)/12

x2 = -12/12

x2 =
-1

The solutions of the equation are: {-1 ;
1/6}.