Trying to find the axis of rotation.
$$ M(x)= \frac{1}{2}\left[ {\begin{array}{cc} -1+\cos\ x & \sqrt{2}\sin\ x & 1+\cos\ x\\ \sqrt{2}\sin\ x & -2\cos\ x & \sqrt{2}\sin\ x\\ 1 + \cos\ x & \sqrt{2}\sin\ x & -1 + \cos\ x \end{array} } \right] $$
I know the trace $= -1$ and that this cooresponds to a rotation of order 2 but I'm not sure how to find the axis of rotation.
One way to compute the rotation axis is to find an eigenvector of $1$, i.e., to compute a basis for the kernel of $M-I$. This involves a slightly messy, but not too difficult row-reduction computation. However, there are a couple of ways to find the rotation axis without referring to eigenvalues or eigenvectors.
As explained here, the simplest way is to take the skew-symmetric part of $M$, which is a scalar multiple of the “cross-product matrix” of a vector on the rotation axis and read off the axis vector coordinates. That is, for a 3-D rotation matrix $R$, $R-R^T=\sin\theta\,K$, where $\theta$ is the rotation angle and $K$ is the axis cross-product matrix. Unfortunately, for the matrix in this question $\operatorname{tr}M=1+2\cos\theta=-1$, which means that $\theta=\pi$ and $M-M^T=0$, which gives us no information about the axis.
Fortunately, with a little more work the axis can also be extracted from the symmetric part of $R$: each of the rows/columns of $T=R+R^T-(\operatorname{tr}R-1)\,I$ consists of the coordinates of a vector on the rotation axis. This method has the advantage over using the skew-symmetric part of working for any rotation angle.
In this case, $$T = \begin{bmatrix} 1+\cos x & \sqrt2 \sin x & 1+\cos x \\ \sqrt2 \sin x & 2-2\cos x & \sqrt2 \sin x \\ 1+\cos x & \sqrt2 \sin x & 1+\cos x \end{bmatrix},$$ and so the rotation axis is $[1+\cos x, \sqrt2 \sin x, 1+\cos x]^T$, which you can verify satisfies $Mv=v$.