I am currently puzzled over how to deal with the following exercise:
$$2\cos(\theta)-1=0$$
Here is the work I have done so far concerning the listed equation:
$$2\cos(\theta)-1=0 \\2\cos(\theta)=1 \\\cos(\theta)=\frac{1}{2} \\\theta=\frac{\pi}{3}+2\pi n \text{ or } \theta=\frac{5\pi}{3}+2\pi n$$
You would think that this is the solution however the obstacle which has stopped me in my tracks is the $k\pi$ with $k\in \mathbb{Z}$. Let me explain. For each of the two solutions which I found, I have added $2\pi n$ since if $2\pi$ is added no matter how many times the point will land on the same place on the unit circle for any value. I think that perhaps the $k\pi$ with $k\in \mathbb{Z}$ is an expression used in mathematics which acts much like a synonym to the $2 \pi n$ which I have added to each solution. I ask of you to inform me more of $k\pi$ with $k\in\mathbb{Z}$ and its use within the listed problem. Thank you ahead of time!
The solution of the problem is, as you said, $\theta=\frac{\pi}{3}+2\pi n \text{ or } \theta=\frac{5\pi}{3}+2\pi n$ with $n\in \mathbb{Z}$ (you can add or subtract $2\pi$ and the cosine doesn't change).
But the solution isn't $\theta=\frac{\pi}{3}+\pi k \text{ or } \theta=\frac{5\pi}{3}+\pi k$ with $k\in \mathbb{Z}$! For example, you can check what happens with $k=1$: if you add (or subtract) $\pi$ then the angle is actually different and the cosine might change. Long story short: the coefficient before the "$k\pi$" is important and you have to write it.
Another example: $\cos \alpha = 0$, then $\alpha = \frac{\pi}{2} + k\pi$. Here it appears the "$k\pi$", and not the "$2k\pi$". This happens because the solutions are "actually" $\alpha=\frac{\pi}{2}+2k\pi \text{ or } \alpha=-\frac{\pi}{2}+2k\pi$, and the other notation (in this case) is equivalent.