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}.**