The short answer to your question is "yes", you can exploit the structure of the Fourier-series to reduce your rootfinding problem to a matrix eigenvalue problem, which can then be solved efficiently using QR decomposition. None of this is my own, but I'm paraphrasing a paper by John P. Boyd at University of Michigan from 2006, entitled "Computing the zeros, maxima and inflection points of Chebyshev, Legendre and Fourier series: solving transcendental equations by spectral interpolation and polynomial rootfinding". Let a finite Fourier-series with $2N+1$ terms be given by
$$
f_N(t) = \sum_{j=0}^N a_j\cos(jt) + \sum_{j=1}^N b_j\sin(jt)
$$
Define the following $2N+1$ coefficients:
$$
h_k = \left\{\begin{array}{ll} a_{N-k}+i b_{N-k}, & k=0,1,\ldots,(N-1)\\
2 a_0 & k=N \\
a_{k-N} - i b_{k-N} & k=N+1,N+2,\ldots, 2N
\end{array}\right.
$$
And define the following $2N\times2N$ matrix $\bf B$ with entries $B_{jk}$ at indicies $j,k$ as
$$
B_{jk} = \left\{
\begin{array}{ll}
\delta_{j,k-1}, & j=1,2,\ldots,(2N-1)\\
- \frac{h_{k-1}}{a_N - i b_{N}} & j=2N
\end{array}\right.
$$
The delta above is the Kronecker delta that is one when its two arguments are the same, zero otherwise. Let the matrix $\bf B$ have eigenvalues given by $z_k$, then the roots of $f_N(t)$ are given by
$$
t_k = -i \log(z_k)
$$
In general $z_k$ will be complex or negative, so we mean the complex logarithm (this nicely gives that each root is also periodic, the expected behavior of a periodic function)
$$
\log(z) = \log(|z|) + i (\arg(z)+2\pi m) \;\;\text{for integer }m
$$
For a final formula for the periodic roots
$$
t_k = \arg(z_k)+2\pi m-i \log(|z_k|),\;\;k=1,2,\ldots,2N,\;\;m\text{ an integer}
$$
So the numerical task of Fourier series rootfinding is reduced to basically solving for eigenvalues of the matrix $\bf B$.