How can I solve $\cos^2 x + \sin x +1 = 0$? The solution set of the equation $$\cos^2 x + \sin x +1 = 0$$ is? I haven't studied trigonometry, I'm kinda lost on this issue ...

We can use the identity $\sin^2 x + \cos^2 x = 1$.
$$\begin{align} \cos^2 x + \sin x + 1 & = 0 \\ \\  (1 - \sin^2 x) + \sin x + 1 & = 0 \\ \\ \sin^2 x - \sin x - 2 & = 0\end{align}$$ 
Now, you have a quadratic in $\sin x$. For example, set $t = \sin x$, and you'll have a quadratic in $t$, find the roots, then find the corresponding solutions to each root.
Putting $\sin x = t,$ we obtain $$t^2 - t - 2 = (t-2)(t+1) = 0$$
Roots: $t = 2,\;\;t = -1$.  
Since $t = \sin x$, that means that 


*

*$t = \sin x = 2$ (impossible, since $-1 \leq \sin x \leq 1, \;\forall x \in \mathbb R$), or 

*$t = \sin x = -1 \implies x = 3\pi/2 + 2k\pi$, where $k$ is any integer.

