An angle $\theta$ can be trisected if and only if $4x^3-3x+\cos\theta$ is reducible over $\mathbb{Q}(\cos\theta)$ I was given the above problem for homework. There is (what seems to be) a relevant proof in my textbook regarding the impossibility of trisecting $\pi/3$. In this proof, the identity
$$\cos 3\theta = 4\cos^3 \theta -3\cos\theta$$
is used. Rearranging, we get $ 0 =  4\cos^3 \theta-3\cos\theta-\cos 3\theta$. I know if the given equation were $4x^3-3x-\cos\theta$, my homework problem would be relatively easy. At this point, however, I'm not sure where to go. A push in the right direction would be very appreciated.
Edit: The question isn't actually out of the textbook (Galois Theory by Stewart). It's on a worksheet my teacher typed up, which makes me think it might be a typo as well. In fact, the textbook asks the analogous question for $4x^3-3x-\cos\theta$.
 A: An angle $\theta$ can be trisected if and only if $4x^3-3x-\cos \theta $ is reducible over $\mathbb Q(\cos \theta)$. It can be proved easily in following manner.
Assume $\phi=\frac{\theta}3$. $\phi$ can be constructed from $\theta$ iff $\cos \phi$ can be constructed from $\cos \theta$. 
Now use $$\cos \theta = \cos 3\phi=4 \cos^3 \phi-3\cos \phi$$
 Hence $\cos \phi$ is a root of $f(x)=4x^3-3x-\cos \theta$.Now if $f(x)$ is reducible over $\mathbb Q(\cos \theta)$,the $\cos \phi$ is a root of polynomial of degree $1$ or $2$  over $\mathbb Q(\cos \theta)$, thus $[\mathbb Q(\cos \theta,\cos \phi):\mathbb Q(\cos \theta)]=1 \ or\ 2$. Hence $\cos \phi$ is constructible from $\mathbb Q(\cos \theta)$.
If $f(x)$ is irreducible over $\mathbb Q(\cos \theta)$, then $[\mathbb Q(\cos \theta,\cos \phi):\mathbb Q(\cos \theta)]=3$ and thus $\cos \phi$ cannot be contructed from $\mathbb Q(\cos \theta)$.
EDIT: As pointed out by Andre Nicholas, $\phi$ is constructible iff $-\phi $ is. Also $4x^3-3x-a$ is irreducible over a field $F$ iff $4x^3-3x+a$ as can be seen by putting change of variable $t=-x$.
