1-4tan(3x+1)=9,
-8=4tan(3x+1),
-2=tan(3x+1),
3x+1=arctan(-2).
Tangent is negative in quadrants II and IV, that is, angles between π/2+2πn and π+2πn, and 3π/2+2πn and 2π+2πn (1.57-3.14+6.28n, 4.71-6.28+6.28n) radians, where n is an integer.
Therefore 3x+1=5.1760+6.2832n or 2.034+6.2832n radians approx.
So 3x=4.1760+6.2832n or 1.034+6.2832n. We need to choose all values of n that keep x in the given range 0-2π, that is, 0-6.2832.
We can write x=1.3920+2.0944n, 0.3448+2.0944n.
When n=0: x=1.3920, 0.3448;
When n=1: x=3.4864, 2.4392;
When n=2: x=5.5808, 4.5336;
We have 6 possible values for x.
The red curves represent the function tan(3x+1) and the blue line represents the value -2. The intersections represent the 6 solutions in the required range of values. These concur approximately with the solution you expected.