Say you plot $ r = \theta $ from [0, 2pi]
Consider an arbitrary angle $\alpha$
The length $r(\alpha)$ can be trisected using a ruler and compass. Arcs can be "swept out" from the points of trisection to the axis.
The smallest arc has radius equal to $\frac{1}{3}r$
Because $r$ is related by $r = \theta$ the angle forming the intersection between $r = \theta$ and the arc of radius $\frac{1}{3}r$ is equal to $\frac{1}{3}\alpha$, allowing us to draw a ray from the origin through the point of intersection which is at an angle $\frac{1}{3}\alpha$.
If you also consider the arc of radius $\frac{2}{3}r$, we now are allowed to also draw a ray from the origin through the point of this intersection which is at an angle $\frac{2}{3}\alpha$. This splits the region swept by $\alpha$ into three separate ones, each of central angle $\frac{1}{3}\alpha$.
Thus $\alpha$ was trisected.
Is this not true?
Your construction is certainly valid.
However, you are overlooking an important part of the classical angle trisection problem:
Your construction is not a ruler and compass construction, because it depends on an object which is itself not constructible by ruler and compass, namely the Archimidean spiral.
So although you may have shown that any angle may be trisected by a "ruler and compass and Archimidean spiral" construction, you have not shown how to do it by a "ruler and compass" construction.
And that is fortunate, because the theorem has been proved that it is not possible to trisect every angle by a ruler and compass construction.