If you extend Willie Wong's comment on the Product-to-sum identity (works better than mine)
$$
\cos \theta_1 \cos \theta_2= {\cos(\theta_1 + \theta_2) + \cos(\theta_1 - \theta_2) \over 2}
$$
(with $\omega_kt+\varphi_k\color{black}{=\theta_k)}$ to $3$ cosines
$$
\begin{eqnarray}
\cos \theta_1 \cos \theta_2 \cos \theta_3&=& {\cos(\theta_1 + \theta_2) + \cos(\theta_1 - \theta_2) \over 2}\cos \theta_3\\
&=&\scriptstyle{\cos(\theta_1 + \theta_2+\theta_3) +\cos(\theta_1 + \theta_2-\theta_3) + \cos(\theta_1 - \theta_2+\theta_3)+\cos(\theta_1 - \theta_2-\theta_3) \over 4},
\end{eqnarray}
$$
you recognize that all $2^{3-1}$ $"\pm"$ combinations of $\theta_k$ appear. So we get
$$
x(t)=\prod_{k=1}^NA_k \cos(\omega_kt+\phi_k)=
\frac A{2^{N-1}}\sum_{b=0}^{2^{N-1}-1}\cos\left(\sum_{k=1}^N (-1)^{b_k}\theta_k\right),
$$
with $\left(\prod_{k=1}^NA_k\right)$ and $b=\sum_{k=1}^{N} 2^{k-1}b_k$. Then use what you already know about the Fourier Transform and you're done.