Step 1: Use the second equation first.
From $3\cos 2\theta = 1$, we get $\cos 2\theta = \dfrac13$.
Step 2: Convert to $\tan^2\theta$.
Using $\cos 2\theta = \dfrac{1-\tan^2\theta}{1+\tan^2\theta} = \dfrac13$: cross-multiplying, $3 - 3\tan^2\theta = 1 + \tan^2\theta$, so $4\tan^2\theta = 2$, giving $\tan^2\theta = \dfrac12$.
Step 3: Compute $32\tan^8\theta$.
$\tan^8\theta = (\tan^2\theta)^4 = \left(\tfrac12\right)^4 = \tfrac{1}{16}$, so $32 \cdot \tfrac{1}{16} = 2$.
Step 4: Reduce the first equation.
The first equation becomes $2\cos^2\alpha - 3\cos\alpha = 2$, i.e. $2\cos^2\alpha - 3\cos\alpha - 2 = 0$.
Step 5: Solve the quadratic.
Factor: $(2\cos\alpha + 1)(\cos\alpha - 2) = 0$. Since $\cos\alpha = 2$ is impossible, $\cos\alpha = -\dfrac12$.
Step 6: Write the general solution.
$\cos\alpha = -\tfrac12 = \cos\tfrac{2\pi}{3}$, so $\alpha = 2n\pi \pm \dfrac{2\pi}{3},\ n\in\mathbb{Z}$, which is option (D).
\[ \boxed{\alpha = 2n\pi \pm \tfrac{2\pi}{3},\ n\in\mathbb{Z}} \]