Trigonometry identities

The given equation is $3(\sec\theta - \tan\theta) = 2\cos\theta$.
We know that $\sec\theta = \frac{1}{\cos\theta}$ and $\tan\theta = \frac{\sin\theta}{\cos\theta}$.
Substitute these into the equation:

$$3\left(\frac{1}{\cos\theta} - \frac{\sin\theta}{\cos\theta}\right) = 2\cos\theta$$

Note that this implies $\cos\theta \neq 0$.

$$3(1 - \sin\theta) = 2\cos^2\theta$$

$$3 - 3\sin\theta = 2 - 2\sin^2\theta$$ $$2\sin^2\theta - 3\sin\theta + 3 - 2 = 0$$ $$2\sin^2\theta - 3\sin\theta + 1 = 0$$

Let $x = \sin\theta$.

$$2x^2 - 3x + 1 = 0$$

Factor the quadratic equation:

$$(2x - 1)(x - 1) = 0$$

This gives two possible solutions for $x$:
$2x - 1 = 0 \implies x = \frac{1}{2}$
$x - 1 = 0 \implies x = 1$

Case 1: $\sin\theta = 1$
The general solution is $\theta = \frac{\pi}{2} + 2n\pi$, where $n$ is an integer.
However, in Step 3, we established that $\cos\theta \neq 0$. If $\sin\theta = 1$, then $\theta = \frac{\pi}{2} + 2n\pi$, which means $\cos\theta = 0$. This would make $\sec\theta$ and $\tan\theta$ undefined in the original equation. Therefore, $\sin\theta = 1$ is an extraneous solution.

Case 2: $\sin\theta = \frac{1}{2}$
The general solutions are $\theta = \frac{\pi}{6} + 2n\pi$ and $\theta = \frac{5\pi}{6} + 2n\pi$, where $n$ is an integer.
For these values of $\theta$, $\cos\theta \neq 0$, so the original terms $\sec\theta$ and $\tan\theta$ are well-defined.

The solutions for $\theta$ are:

$$\boxed{\theta = \frac{\pi}{6} + 2n\pi \quad \text{or} \quad \theta = \frac{5\pi}{6} + 2n\pi, \quad \text{where } n \in \mathbb{Z}}$$