<s>[INST] For which values of $k$ does the equation $2\cos^{2}\theta +k\sin \theta + k = 2$ have real solutions? So I take A level maths and this question was in our textbook. We solved an inequality for when the discriminant is less than zero and this gave us  the same answer that is in the textbook. The problem is that the solutions also have to give $\sin \theta$ as between one and minus one, and we didn't know how to solve that inequality.
Any help would be appreciated, thanks!
 [/INST] Rewriting $2\cos^2\theta$ as $2-2\sin^2\theta$, we can substitute $t = \sin\theta$ and our equation becomes
$$
2t^2-kt-k = 0
$$
with solutions
$$
t = \frac{k\pm\sqrt{k^2+8k}}{4}
$$
Having a non-negative discriminant $k^2+8k$ forces either $k \leq -8$ or $k \geq 0$.  But in addition, we require $-1 \leq t \leq 1$ for at least one of the solutions, so that $\theta$ can be real.  This is equivalent to
$$
-4 \leq k\pm\sqrt{k^2+8k} \leq 4
$$
for at least one of the branches of the square root.
Case $1$ ($k \leq -8$).  It is clear that $k - \sqrt{k^2+8k} < -4$, so we focus on $k + \sqrt{k^2+8k}$.   But then
\begin{align}
k+\sqrt{k^2+8k} & < k+\sqrt{k^2+8k+16} \\
                & = k+(-k-4) \\
                & = -4
\end{align}
so Case $1$ yields no real solutions.
Case $2$ ($k \geq 0$).  We see that $k - \sqrt{k^2+8k} \leq 0$.  When is it at least $-4$?
\begin{align}
k - \sqrt{k^2+8k} & > k - \sqrt{k^2+8k+16} \\
                  & = k - (k+4) \\
                  & = -4
\end{align}
Hence there exists (at least) one real solution for all $k \geq 0$.  The other solution is always non-negative but exceeds $1$ when
$$
k + \sqrt{k^2+8k} > 4 \\
4-k < \sqrt{k^2+8k} \\
k^2-8k+16 < k^2+8k \\
16k > 16 \\
k > 1
$$
Note: For $k > 4$, we can no longer go from the second line to the third line, but then $k + \sqrt{k^2+8k} > k > 4$ anyway.
</s>