Sometimes its easier to work by inspection, noting any obvious properties a matrix may have, than to proceed with the full formalism of the characteristic polynomial, etc.; I think such is the case with the present matrix. Setting
$A = \begin{bmatrix} \cos \theta & 0 & -\sin \theta \\ 0 & 1 & 0 \\ \sin \theta & 0 & \cos \theta \end{bmatrix}, \tag{1}$
it is very easy to see that, with
$\mathbf j = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}, \tag{2}$
we have
$A \mathbf j = \mathbf j = 1 \mathbf j, \tag{3}$
which shows us that $1$ is an eigenvalue of $A$ with eigenvector $\mathbf j$. Next we observe that, setting
$\mathbf i = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \tag{4}$
and
$\mathbf k = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}, \tag{5}$
we have
$A \mathbf i= \cos \theta \mathbf i + \sin \theta \mathbf k \tag{6}$
and
$A\mathbf k = -\sin \theta \mathbf i + \cos \theta \mathbf k. \tag{7}$
(6) and (7) together show that the subspace $\mathcal S = \text{span}\{ \mathbf i, \mathbf k \}$ is invariant under the action of $A$; that is $A \mathcal S \subset \mathcal S$. Therefore any eigenvalue of the restriction of $A$ to $\mathcal S$, $A_{\mathcal S}$, will necessarily be an eigenvalue of $A$. It is easy to see that the matrix $A_{\mathcal S}$ is
$A_{\mathcal S} = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}, \tag{8}$
and the characteristic polynomial $p_{A_{\mathcal S}}(\lambda)$ of $A_{\mathcal S}$ is given by the simple quadratic
$p_{A_{\mathcal S}}(\lambda) = \det(A_{\mathcal S} - \lambda I) = (\cos \theta - \lambda)^2 + \sin^2 \theta$
$= \lambda^2 - 2(\cos \theta) \lambda + \cos^2 \theta + \sin^2 \theta = \lambda^2 - 2(\cos \theta)\lambda + 1; \tag{9}$
it is easy to see from the quadratic formula that the roots of $p_{A_{\mathcal S}}(\lambda)$ are
$\lambda = \dfrac{1}{2}(2\cos \theta \pm \sqrt{4\cos^2 \theta - 4}) = \cos \theta \pm \sqrt{-\sin^2 \theta} = \cos \theta \pm i \sin \theta = e^{\pm i \theta}; \tag{10}$
we thus see these eigenvalues are in agreement with those given by Victor Liu in his answer.
In the above calculations, we have effectively replaced factoring the cubic characteristic equation of $A$ with the decomposition of $\Bbb R^3$ (or $\Bbb C^3$) into two separate invariant subspaces of $A$, $\text{span}\{\mathbf j \}$ and $\text{span} \{ \mathbf i, \mathbf k \}$.
Of course one can if one chooses exploit the cubic $p_A(\lambda)$ directly:
$p_A(\lambda) = \det (A - \lambda I) = \det(\begin{bmatrix} \cos \theta - \lambda & 0 & -\sin \theta \\ 0 & 1 - \lambda & 0 \\ \sin \theta & 0 & \cos \theta - \lambda \end{bmatrix})$
$ = (\cos \theta - \lambda)^2(1 - \lambda) + \sin^2 \theta (1 - \lambda) = ((\cos \theta - \lambda)^2 + \sin^2 \theta)(1 - \lambda)$
$= (\lambda^2 - 2(\cos \theta)\lambda + 1)(1 - \lambda), \tag{11}$
where evaluation of the determinant is made easy using Sarrus' rule. It is easy to see from (11) that the roots of $p_A(\lambda)$ are the same as the eigenvalues discovered above, basically by "inspection".
Hope this helps. Cheers,
and as always,
Fiat Lux!!!
Best Answer
The matrix has rank $2$, so the nullspace has dimension $3$. That means $\lambda=0$ is an eigenvalue with multiplicity at least $3$.
Since the matrix is symmetric, it is diagonalizable, so it cannot be that all eigenvalues are equal to $0$.
(Alternatively, since $A^2\neq 0$ (because $A^2(0,1,0,0,0) = A(1,0,0,0,0)\neq (0,0,0,0,0)$) then the characteristic polynomial cannot be $-t^5$ (the Jordan form would have at least three blocks, so the largest block has size at most $2$, and then $A^2$ would be zero).
The trace is $0$, so that means that the other two eigenvalues are of the form $\lambda$ and $-\lambda$ for some $\lambda\neq 0$.
If $(a,b,c,d,e)$ is an eigenvector of $\lambda$, with $\lambda\neq 0$, then $\lambda a=b+c+d+e$, and $a=\lambda b=\lambda c=\lambda d=\lambda e$. So $b=c=d=e = \frac{a}{\lambda}$, which gives $\lambda^2=4$. Thus, if there are any nonzero eigenvalues, then they should be $\lambda=2$ and $\lambda=-2$. So the eigenvalues are $0$ (with multiplicity $3$), $2$, and $-2$.