Solving $-1\leq \frac{-k \pm \sqrt{{k^2}-8k}}{4}\leq 1.$ $$-1\leq \frac{-k \pm \sqrt{{k^2}-8k}}{4}\leq 1$$
The problem is the square root, as I don't think I can just square both sides. Any ideas?
So this actually comes from a quadratic in $$\sin{\theta}$$
Thanks.
 A: The middle term is actually the two roots $x_{\pm}$ [not necessarily distinct] for the equation
$$
2x^2 + kx + k = 0. 
$$
Let $f(x) = 2x^2 + kx + k$. Then this is an upward parabola symmetric w.r.t. the line $x = -k/4$. Note that it always passes through the point $(-1, 2)$. Then by the graph of this function, $x_{\pm } \in [-1, 1]$ iff
$$
\begin{cases}
\Delta = k^2 - 8k \geqslant 0, \\
-\dfrac k 4 \in [-1, 1], \\
f(-1) \geqslant 0, \\
f(1) \geqslant 0, 
\end{cases} \iff 
\begin{cases}
k \in (-\infty, 0] \cup [8, +\infty), \\
k \in [- 4,4] \\
2 \geqslant 0, \\
2 + 2k \geqslant 0, 
\end{cases}
$$
iff
$$
\boxed {\boldsymbol {k \in [-1, 0]}}\ .
$$
Update
If it means either one lies in the interval $[-1,1]$, then
$$
\begin{cases}
\Delta = k^2 - 8k \geqslant 0, \\
-\dfrac k 4 \geqslant -1, \\
f(1) \leqslant 0,
\end{cases}
$$
also works, and the final result is $\boldsymbol {(-\infty, 0]}$ instead.
A: Write it as
$$ k - 4 \le \pm \sqrt{k^2 - 8k} \le k + 4$$
Now there will be cases depending on whether $k$ is in the interval $(-\infty, -4]$, $[-4,4]$ or $[4,\infty)$.
EDIT: For example, if $k \in (-\infty, -4]$, the $\pm$ must be $-$, and
$k-4 \le -\sqrt{k^2 - 8 k} \le k+4 \le 0$ becomes
$$0 \le (k+4)^2 \le k^2-8k \le (k-4)^2$$
Expand the squares and subtract $k^2-8k$...
