Find the points on the curve y=(sinx)/(2+cosx) at which the tangent is horizontal. Supplemental question about a different scenario. 
Find the points on the curve $ y = \frac { \sin x }{ 2 + \cos x} $ at which the tangent is horizontal.

I'd like some input on whether I solved this correctly or not as trig has never been a particularly strong point for me. I found a similar question and moved forward based off reading through this thread. 
$$y'=\frac{2\cos x + 1}{(2+\cos x)^2}$$
I'm not exactly sure why, but from the previously linked thread, it looks like I shouldn't be concerned about the $ (2 + \cos x)^2 $ because it cannot equal zero, so I can just put $ 2 \cos x = 0$ and solve so $ x = \frac{2\pi}{3} , x = \frac{5\pi}{3}$ then plugging it back into the original equation I end up with $ y (\frac{2\pi}{3}) = \frac{\sqrt3}{3}$ and $ y (\frac{5\pi}{3}) = -\frac{\sqrt3}{3}$
Therefore, there exist horizontal tangents at $ ( \frac{2\pi}{3} + 2\pi n , \frac{\sqrt3}{3} ),( \frac{5\pi}{3} + 2\pi n , -\frac{\sqrt3}{3} ), n\in\mathbb{Z}$
As some supplemental inquiries, how would the results be altered if the equation were this?$$ \frac{ \sin x }{ 1 + \cos x} $$ There would be a vertical aymptote when $ x = \pi $ right? Would that affect the horizontal tangent at all? And could I get an explanation as to why I don't need to put the whole equation equal to zero?
Edit: Altered answer to include for infinite values where the derivative is equal to zero and correct an error
 A: 
I'm not exactly sure why, but from the previously linked thread, it looks like I shouldn't be concerned about the $(2+\cos x)^2$ because it cannot equal zero

This is because a fraction can only be zero when the numerator is zero ($\frac{1}{n}$ is never zero no matter what $n$ is, similarly $\frac{x}{n}$ is only zero when $x$ is zero). However, you do need to check if the fraction is defined (the denominator is not zero). In this case it will never be zero, because $\cos x \neq -2$.
Now the numerator is zero when 
$$2\cos x + 1 = 0 \,\, \Rightarrow x=\frac{2}{3}(3\pi n-\pi), n\in\mathbb{Z}$$
The derivative has an infinite number of zeros, meaning there's an infinite number of points of horizontal tangency. 
In this case $y(x)$ is bounded and $y(\frac{2}{3}(3\pi n-\pi)) = \pm\frac{\sqrt{3}}{3}, n\in\mathbb{Z}$ so all lines of horizontal tangency lie on the lines $y=\pm\frac{\sqrt{3}}{3}$

how would the results be altered if the equation were $\frac{\sin x}{1+\cos x}$ ?

Your intuition is correct, there will a vertical asymptote at $x=\pi$ and in fact at $x=\pi (2n+1), n\in\mathbb{Z}$. However, $\frac{d}{dx}(\frac{\sin x}{1+\cos x}) = \frac{1}{1+ \cos x}$, which like $\frac{1}{n}$ can never be zero.
