How do I compute the fourier transform of a signal:
$$Ax(t)\cos(\omega t) $$
where both $A$ and $\omega$ are constants. I tried using the fact the function was even and using Euler's formula but I ended up with $2$ equations, both of which equal zero.
The convolution theorem reads $$ \sqrt{2\pi} \mathcal{F} ( f g) =\mathcal{F}(f) * \mathcal{F}(g) $$ for any functions $f(t)$, $g(t)$ and where $$ (F*G)(\nu) = \int d\mu F(\nu-\mu) G(\mu) = \int d\mu F(\mu) G(\nu-\mu) .$$ Apply this to the functions $$f(t) = x(t)$$ and $$g(t) = A \cos(\omega t).$$
Let us denote the Fourier transform of $f(t)$ by $F(\nu)= X(\nu)$. The Fourier transform of $g(t)$ can be explicitly obtained as $$G(\nu) = \frac{\sqrt{2\pi}A}{2}\left[ \delta(\nu-\omega) + \delta(\nu+\omega)\right]\;.$$
Performing the convolution, we obtain that $$\mathcal{F}[A x(t) \cos(\omega t)](\nu)=\frac{A}{2}[X(\nu-\omega)+X(\nu+\omega)].$$ We see that the carrier frequency $\omega$ shifts the frequencies $\nu$ of the amplitude modulation to $\nu \pm \omega$.