Roots of trigonometric equation $1 + \alpha^2 \cos^2 (n \theta) = 0$

Question

In the following trigonometric equation

$$1 + \alpha^2 \cos^2 (n \theta) = 0$$

The complex solutions are

$$\cos (n \theta) = \pm i/\alpha$$

So I thought that the corresponding angles were,

$$n \theta = \arccos (i/\alpha) + 2k\pi$$

$$n \theta = 2\pi - \arccos (i/\alpha) + 2k\pi$$

and,

$$n \theta = \arccos (-i/\alpha) + 2k\pi$$

$$n \theta = 2\pi - \arccos (-i/\alpha) + 2k\pi$$

as usual in trigonometric equations containing cosine. But instead the solutions should be:

$$n \theta = \arccos (i/\alpha) + k\pi$$

and,

$$n \theta = \arccos (-i/\alpha) + k\pi$$

Why?

Answer 1

Note that $\arccos(z) = \pi - \arccos(-z)$.
So $$\arccos(i/\alpha) + (2k-1)\pi = - \arccos(-i/\alpha) + 2k\pi$$ and $$\arccos(-i/\alpha) + (2k-1)\pi = - \arccos(i/\alpha) + 2k\pi$$ Thus these are just two different ways of writing the same set of solutions.