You need to use the following trigonometric identities, such that:

`tan 2x = (2tan x)/(1 + tan^2 x)`

`tan 3x = tan (2x + x) = (tan 2x + tan x)/(1 + tan 2x*tan x) `

`tan 3x = ((2tan x + tan x + tan^3 x)/(1 + tan^2 x))/((1 + 3tan^2 x)/(1 + tan^2 x))`

Reducing duplicate terms yields:

`tan 3x = (tan x(2tan x + tan x + tan^3 x))/(1 + 3tan^2 x)`

`tan 3x = (tan x(3 + tan^2 x))/(1 + 3tan^2 x)`

Substituting `(2tan x)/(1 + tan^2 x)` for `tan 2x` and `(tan x(3 + tan^2 x))/(1 + 3tan^2 x)` for `tan 3x` yields:

`4(tan x(3 + tan^2 x))/(1 + 3tan^2 x) + (8tan x)/(1 + tan^2 x) + 5 tan x = 0`

Factoring out `tan x` yields:

`tan x ((4(3 + tan^2 x))/(1 + 3tan^2 x) + 8/(1 + tan^2 x) + 5) = 0`

You need to solve for x the equations `tan x = 0` and `((4(3 + tan^2 x))/(1 + 3tan^2 x) + 8/(1 + tan^2 x) + 5) = 0` , such that:

`tan x = 0 => x = arctan 0 + n*pi => x = n*pi, n in Z`

`((4(3 + tan^2 x))/(1 + 3tan^2 x) + 8/(1 + tan^2 x) + 5) = 0`

`4(3 + tan^2 x)(1 + tan^2 x) + 8(1 + 3tan^2 x) + 5(1 + 3tan^2 x)(1 + tan^2 x) = 0`

`12 + 16tan^2 x + 4tan^4 x + 8 + 24tan^2 x + 5 + 20tan^2 x + 15tan^4 x = 0`

`19tan^4 x + 60tan^2 x + 25 = 0`

You may use the following substitution, such that:

`tan^2 x = t`

`19t^2 + 60t + 25 = 0`

Using quadratic formula yields:

`t_(1,2) = (-60 +- sqrt(3600 - 1900))/38`

`t_(1,2) = (-60 +- sqrt(1700))/38`

Since `t_(1,2) < 0` , hence, there exists no value for x such that `tan^2 x = t.`

**Hence, evaluating the solutions to the given trigonometric equation
yields `x = n*pi, n in Z` .**

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