Trisection of an angle $ Z $ whose cosine is $ -\frac{11}{16} $ with straightedge and a compass

Question

Suppose there exists an angle $Z$ such that $\cos Z = -\frac{11}{16}$. Prove or disprove that such an angle can be trisected with a straightedge and a compass.

Well, we know that an angle is constructible if and only if its cosine is constructible. Therefore, we have to prove that $\cos (\frac{Z}{3})$ is constructible... However, I don't know what to do next.

Answer 1

From the addition theorems, find the cubic equation for $\cos\frac13 Z$. Either it has a rational root or it is irreducible. In the latter case $\cos\frac13Z$ is not constructible (because $3$ is not a power of $2$).

Answer 2

Step 1:

$$\cos(3z)=4\cos^3z-3\cos z$$

Step 2:

$$4c^3-3c+\frac{11}{16}=\frac1{16}(4c-1)(16c^2+4c-11)$$