You need to remember that the tangent function is rational, hence `tan 2x = (sin 2x)/(cos 2x).`

The problem provides the information that the angle x is in quadrant 3, hence `sin xlt0 ; cos xlt 0` .

You need to remember the formula of half of angle such that:

`sin x = sqrt((1-cos 2x)/2)`

Substituting `-1/3` for sin x yields:

`-1/3 = sqrt((1-cos 2x)/2)`

You need to raise to square to remove the square root such that:

`1/9 = (1-cos 2x)/2 =gt 2/9 = 1 - cos 2x`

`cos 2x = 1 - 2/9 =gt cos 2x = 7/9`

You need to use the basic formula of trigonometry to find sin 2x such that:

`sin 2x = sqrt(1 - cos^2 2x)`

`sin 2x = sqrt(1 - 49/81) =gt sin 2x = sqrt(32/81)`

`sin 2x = sqrt32/9`

You need to substitute `sqrt32/9` for `sin 2x` and `7/9` for `cos 2x` in `tan 2x = (sin 2x)/(cos 2x) ` such that:

`tan 2x = (sqrt32/9)/(7/9) =gt tan 2x = (sqrt32/9)*(9/7)`

`tan 2x = sqrt32/7`

**Hence, evaluating the tangent of double of angle x yields `tan 2x = sqrt32/7.` **

We'll apply the double angle identity for tan (2x):

tan (2x) = 2tan x/[1-(tan x)^2] If x>=pi and x=<3pi/2, then x is located in the 3rd quadrant, where the values of tangent function are positive.

tan x = sin x/cos x

We need to determine cos x from Pythagorean identity:

cos x = -sqrt[1-(sin x)^2]

cos x = -sqrt(1 - 1/9)

cos x = -sqrt (8/9)

Since x is in the 3rd quadrant, the values of cosine function are negative.

tan x = (-1/3)/(-2sqrt2/3)

tan x = 1/2sqrt2

tanx = sqrt2/4

tan (2x) = 2* (sqrt2/4)/(1-1/8)

tan (2x) = [(sqrt2)/2]/(7/8)

tan (2x) = 4sqrt2/7

**The requested value of tan(2x) = 4sqrt2/7.**