Let $x$ is the angle of the diagonal lines of a unit cube, then what are the possible values of $\cos x$? 
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*Let $x$ is the angle of the diagonal lines of a unit cube(the cube with edges $1$), then what are the possible values of $\cos x$?

My attempt: Since the dot products of the diagonal lines are $1$ or $-1$,and the lengths of them are all $3^{1/2}$, so $\cos x$ are
$$1/(3^{1/2}3^{1/2})=\cos x=1/3$$ or $$-1/(3^{1/2}3^{1/2})=\cos x=-1/3.$$
Am I right? If I am wrong, then please correct my answers.


*Moreover, it should have $16$ possible $\cos x$ values, right?

 A: In fact, you can have a clearer view by using coordinates, with vertices $(\pm 1, \pm 1, \pm 1)$, the cube's center $O$ being the origin of coordinates (see figure below).
Remark: instead of a cube with unit sides, we have a cube with sidelength $2$. We compensate it by taking the half-diagonals, finally giving the same results.
If we considering the different cases:
$$\vec{OA}\begin{pmatrix}\ \ 1\\ \ \ 1\\-1\end{pmatrix}, \vec{OB}\begin{pmatrix}-1\\ \ \ 1\\-1\end{pmatrix},\cdots \vec{OF}\begin{pmatrix}-1\\ \ \ 1\\ \ \ 1\end{pmatrix},\cdots$$
we find that, using formula $\vec{u}.\vec{v}=\frac{u_x*v_x+u_y*v_y+u_z*v_z}{\|\vec{u}\|*|\vec{v}\|}$:
$$\vec{OA}.\vec{OB}=\frac{1*(-1)+1*1+(-1)*(-1)}{\sqrt{3}\sqrt{3}}=\frac13$$
whereas:
$$\vec{OA}.\vec{OF}=\frac{1*(-1)+1*1+(-1)*1}{\sqrt{3}\sqrt{3}}=-\frac13$$
and we have only this kind of dot-product as you have found (one must of course eliminate the cases where the dot product is $-1$, because it corresponds to opposite half-diagonals), giving finally two kinds of angles only, that can be either
$$\cos^{-1}\tfrac13\approx 70.53°, \ \ \text{or} \ \ \cos^{-1}(-\tfrac13)\approx 109.47°  $$
This last angle, $109.47°$ is well know by chemists : it is the "bond angle" in $CH_4$, the methane tetrahedric molecule.
I have materialized (black lines) the way a regular tetrahedron $BDEG$ is inscribed in a cube.
(the following Geogebra figure can be "animated" : click here for that).

