Can one trisect the angle $\theta=\arccos(-12/17)$? I think this is not possible, and below is my proof.
Proof: 
Using the angle tripling formula:
$$\cos(3\theta) = 4\cos^3 (\theta)−3\cos(\theta) \ \to \ \cos(\theta) = 4\cos^3 (\theta/3) -3\cos(\theta/3)$$  here $\cos(\theta) =-12/17$ so we have $4\cos^3 (\theta) -3\cos(\theta) =-12/17$. This implies $$68x^3-51x+12=0,\  x=\cos(\theta/3)$$
Hence $f(x)=68x^3-51x+12$ is a polynomial with root $\cos(\theta/3)$. Notice that $f(x)$ is the minimal polynomial with this root by Eisenstein's criterion with $p=3$. Hence, $[\mathbb{Q}[\cos(\theta/3) ]:\mathbb{Q}]=3$ which is not a power of two.
 A: Having solved the problem at hand, let us explore a more general result.
Theorem.  Suppose an angle $3\theta$ is given as the inverse cosine of a rational fraction expressed in lowest terms.  Then the trisected angle $\theta$ is constructible only if the denominator is a power of 2 times a cube.
Let $m/n$ be the given value of $\cos3\theta$ with $m,n$ relatively prime.  Then the cubic equation
$n\cos3\theta-m=4n\cos^3\theta-3n\cos\theta-m=0$
has constructible roots only if it has a rational root $p/q$ with $p,q$ relatively prime.  Then we must have
$\dfrac{m}{n}=4(\dfrac{p}{q})^3-3(\dfrac{p}{q})=\dfrac{4p^3-3pq^2}{q^3}$
Then $m/n$ in lowest terms implies
$n=q^3/\text{gcd}(4p^3-pq^2,q^3)$
Since $p$ is prime to $q$, the numerator cannot have any odd prime factors in common with $q$, so the quotient must have the same largest odd divisor as $q^3$.  The claimed result follows.
So $\cos3\theta=-12/17$ cannot be solved for $\theta$ by construction because $17$ is odd and not a cube.
The reader might note that the numerator is a multiple of $3$ in this problem.  If that's the case and the trisection is to be possible, then the reader is invited to prove that the numerator would also have to be a multiple of $9$.  No dice again for $\cos3\theta=-12/17$.
