Let $V$ be the vector space of $3 \times 3$ real symmetric matrices of trace $0$.
It has an obvious $5$-element basis.
$\operatorname{SO}(3)$ acts on $V$ by conjugation $g.v \mapsto g^{-1}vg \in V$.
The character of this representation of $\operatorname{SO}(3)$, $\chi_v$, can be calculated on the maximal torus of $\operatorname{SO}(3)$. By direct calculation, simply multiplying matrices, one has that $\chi_v(t) = 2\operatorname{cos}(2t) + 2\operatorname{cos}(t) + 1$.
One way to make this calculation quicker is to focus on certain coordinates. For instance for the basis element
$$e_1 := \begin{bmatrix} 1 & 0 & 0\\ 0 & -1 & 0\\ 0 & 0 & 0 \end{bmatrix},$$
one can simply focus on the fist coordinate of a matrix $g^{-1}e_1g$ to find out the coefficient of $e_1$ in a basis representation. However, this is still quite tedious and long.
Any way to make a quicker calculation?
(I thought of also trying to represent this vector space representation as irreps of $\operatorname{SU}(2)$, but in order to find out how to write it as one you need to multiply these matrices as far as I can tell, so may as well just find the character).
Let $W$ be the 3-dimension irrep of $SO(3)$ (hence also of $SU(2)$). Then the space of all $3\times 3$ matrices is the representation $W^*\otimes W$. But $W^*\cong W$ so this is $W^{\otimes 2}$. We split off $\Lambda^2 W\cong W$ and the trivial representation (the trace), so what is left is $$ \underbrace{(e^{it}+1+e^{-it})^2}_{W^{\otimes 2}}-\underbrace{(e^{it}+1+e^{-it})}_{W}-\underbrace{1}_{\text{trivial}}=e^{2it}+e^{it}+1+e^{-it}+e^{-2it}. $$