So when trying to compute the Fourier transform I believe I can use the convolution theorem to evaluate this as a whole.
$$ \widehat{g*h}(x) = 2\pi \hat{g}(k) \cdot \hat{h}(k) $$
If it let $ g(x) = \cos(x) $ and $ h(x)= f(2x+3) $, I can find the Fourier transform of $g(x)$ to be something like
$$ \hat{g}(k) = \pi \delta(k-k_{0})+\pi \delta(k+k_{0}) $$
But i'm stuck on how to evaluate the $\hat{h}(k)$ portion.
$\hat{f}(2x + 3) = \int_{-\infty}^\infty e^{- i \omega x} f(2 x+3) dx$
Now, $u=2 x + 3$ gives $du = 2 dx$ and $x = \frac{u-3}{2}$. Plugging this into the integral, you get
$\hat{f}(2x + 3) = \frac{1}{2} \int_{-\infty}^\infty e^{- i \omega (\frac{u-3}{2}) } f(u) du = \frac{1}{2} \int_{-\infty}^\infty e^{- i \frac{\omega}{2} (u-3) } f(u) du = \frac{1}{2} e^{-i \frac{\omega}{2} (-3)} \int_{-\infty}^\infty e^{- i \frac{\omega}{2} u } f(u) du = \frac{1}{2} e^{i \frac{3\omega}{2}} \hat{f}(\omega/2)$
Adjust for your definition of the FT appropriately.
Alternatively, combine the rules for fourier transforms:
$f(ax) \Leftrightarrow \frac{1}{|a|}\hat{f}(\omega/a)$
$f(x-a) \Leftrightarrow e^{-i a \omega} \hat{f}(\omega)$
(using the appropriate form for your version of the FT)