Step 1: Group the terms smartly.
Take the equation $1 + \cos 2\theta + \cos 4\theta + \cos 6\theta = 0$ and pair the first two and the last two.
Step 2: Simplify the first pair.
Using $1 + \cos 2\theta = 2\cos^2\theta$, the first pair becomes $2\cos^2\theta$.
Step 3: Simplify the second pair.
Using $\cos 4\theta + \cos 6\theta = 2\cos 5\theta \cos\theta$, the second pair becomes $2\cos 5\theta \cos\theta$.
Step 4: Factor out the common piece.
The equation is now $2\cos\theta(\cos\theta + \cos 5\theta) = 0$. Combine inside: $\cos\theta + \cos 5\theta = 2\cos 3\theta \cos 2\theta$. So we get $4\cos\theta \cos 2\theta \cos 3\theta = 0$.
Step 5: Solve each factor in $[0^\circ, 180^\circ]$.
$\cos\theta = 0$ gives $\theta = 90^\circ$. $\cos 2\theta = 0$ gives $2\theta = 90^\circ, 270^\circ$, so $\theta = 45^\circ, 135^\circ$. $\cos 3\theta = 0$ gives $3\theta = 90^\circ, 270^\circ, 450^\circ$, so $\theta = 30^\circ, 90^\circ, 150^\circ$.
Step 6: Collect all distinct solutions.
Putting them together: $\theta = 30^\circ, 45^\circ, 90^\circ, 135^\circ, 150^\circ$. \[ \boxed{\theta = 30^\circ, 45^\circ, 90^\circ, 135^\circ, 150^\circ} \]