<s>[INST] How to show that $ 0< a\leq\cos^2(\theta)\leq b<1$ in this problem? The inequality $2\cos^4(\theta/2)-2\cos^2(\theta/2)+1/4\leq 0$  means that $\cos^2(\theta/2)$ lies between the roots of $2x^2-2x+1/4$ i. e., we can conclude that 
$$
\frac{2-\sqrt{2}}{4}\leq\cos^2(\theta/2)\leq\frac{2+\sqrt{2}}{4}.
$$
The question is: from this, can we conclude that
$$
0< a\leq\cos^2(\theta)\leq b<1
$$
for some numbers $a$ and $b$ or, at least $0< a\leq\cos^2(\theta)$?
 [/INST] You can use the "double angle formula"
$$
\cos(\theta) = 2\cos^2(\theta / 2) - 1
$$
so
$$
\cos^2(\theta) = (2\cos^2(\theta / 2) - 1)^2
$$
Then we write
\begin{align*}
\frac{2-\sqrt{2}}{4} &\leq \cos^2(\theta/2)\leq\frac{2+\sqrt{2}}{4} \\
\frac{2-\sqrt{2}}{2} &\leq 2\cos^2(\theta/2)\leq\frac{2+\sqrt{2}}{2} \\
\frac{-\sqrt{2}}{2} &\leq 2\cos^2(\theta/2) - 1\leq\frac{\sqrt{2}}{2} \\
\end{align*}
which implies
\begin{align*}
0 &\leq (2\cos^2(\theta/2) - 1) ^2 \leq \left( \frac{\sqrt{2}}{2} \right)^2 = \frac{1}{2} \\
0 &\leq \cos(\theta) ^2 \leq \frac{1}{2} \\
\end{align*}
You can't do better than $0$ for a lower bound, though, because it is possible that $\theta = \frac{\pi}{2}$.  Then $\cos(\theta) = 0$, and $\theta$ satisfied the required inequality
$2\cos^4(\theta/2)-2\cos^2(\theta/2)+1/4\leq 0$.
</s>