Hybrid Sampling Scheme Idea?
i will reframe the question a little:

we have

$$\begin{align}
 x(t) &amp;= \sum\limits_{n=1}^N r_n \cos(\omega_n t + \phi_n) \ + \ \epsilon(t) \\ 
\\
  &amp;= \sum\limits_{n=1}^N a_n \cos(\omega_n t) + b_n \sin(\omega_n t) \ + \ \epsilon(t) \\ 
\end{align}$$

where 

$$\begin{align}
 a_n &amp;= r_n \cos(\phi_n) \\ 
 b_n &amp;= -r_n \sin(\phi_n) \\ 
\\
 r_n &amp;= \sqrt{a_n^2 + b_n^2} \\
\\
 \phi_n &amp;= \arg\{a_n - j b_n \} \\ 
        &amp;= -\operatorname{atan2}(b_n,a_n) \\
\end{align}$$

and $\epsilon(t)$ is the unknown error signal.

first you have to assume that $\epsilon(t)$ contains no components having frequencies of any of the $\omega_n$ in it, otherwise that component will just team up with the component having amplitude $r_n$.

if you know all of the $\omega_n$ in advance, you can calculate all of the $a_n, b_n$ by using an inner product:

$$\begin{align}
a_n &amp;= \frac{\omega_n}{M\pi} \int\limits_{t_0}^{t_0+\tfrac{2M\pi}{\omega_n}} x(t) \cos(\omega_n t) \ dt \\
\\
b_n &amp;= \frac{\omega_n}{M\pi} \int\limits_{t_0}^{t_0+\tfrac{2M\pi}{\omega_n}} x(t) \sin(\omega_n t) \ dt \\
\end{align}$$

$t_0$ can be the most convenient time for you and $M$ should be a large positive integer (and need not be the same $M$ for each frequency component $\omega_n$), but the wider your integration is, the better error rejection you will have.

from the $a_n$ and $b_n$ coefficients, you can calculate $r_n$ and $\phi_n$.