Given the equation 2 3theta − 1 = 0.

Given the equation $2 \cos 3\theta - 1 = 0$.
Add 1 to both sides:

$$2 \cos 3\theta = 1$$

Divide by 2:

$$\cos 3\theta = \frac{1}{2}$$

Let $\alpha$ be the principal value such that $\cos \alpha = \frac{1}{2}$.

$$\alpha = \arccos\left(\frac{1}{2}\right)$$ $$\alpha = 60^\circ$$

The range for $\theta$ is $0^\circ \le \theta \le 360^\circ$.
Therefore, the range for $3\theta$ is $3 \times 0^\circ \le 3\theta \le 3 \times 360^\circ$, which is $0^\circ \le 3\theta \le 1080^\circ$.

Since $\cos 3\theta$ is positive, $3\theta$ lies in the first and fourth quadrants.
The general solutions for $\cos x = \cos \alpha$ are $x = 2n\pi \pm \alpha$ (in radians) or $x = n \cdot 360^\circ \pm \alpha$ (in degrees), where $n$ is an integer.

For $n=0$:
$3\theta = 60^\circ$
$3\theta = 360^\circ - 60^\circ = 300^\circ$

For $n=1$:
$3\theta = 360^\circ + 60^\circ = 420^\circ$
$3\theta = 360^\circ + (360^\circ - 60^\circ) = 360^\circ + 300^\circ = 660^\circ$

For $n=2$:
$3\theta = 2 \cdot 360^\circ + 60^\circ = 720^\circ + 60^\circ = 780^\circ$
$3\theta = 2 \cdot 360^\circ + (360^\circ - 60^\circ) = 720^\circ + 300^\circ = 1020^\circ$

All these values are within the range $0^\circ \le 3\theta \le 1080^\circ$.

$$\theta_1 = \frac{60^\circ}{3} = 20^\circ$$ $$\theta_2 = \frac{300^\circ}{3} = 100^\circ$$ $$\theta_3 = \frac{420^\circ}{3} = 140^\circ$$ $$\theta_4 = \frac{660^\circ}{3} = 220^\circ$$ $$\theta_5 = \frac{780^\circ}{3} = 260^\circ$$ $$\theta_6 = \frac{1020^\circ}{3} = 340^\circ$$

The values of $\theta$ are $\boxed{20^\circ, 100^\circ, 140^\circ, 220^\circ, 260^\circ, 340^\circ}$.