Trig double angle question

tan 3x = 0 in the interval of [-pi, 0]
How do you solve it? do you use the double angle formula for tan? tan2x= 2tanx/(1-tan^2 x)
I'm really confused!
Thanks so much for helping out!

Why would you use a double angle formula for a triple angle? Just solve for 3x on [-3pi, 0] then divide by 3

3x = 0, -pi, -2pi, -3pi

x = 0, -pi/3, -2pi/3, -pi

Substitute

3x = t

tan(t) = 0
t = pi * k
3x = pi * k
x = (pi/3) * k
x = -pi , (-2/3) * pi , (-1/3) * pi , 0

EDIT:

k is any integer