I have done the following:
\begin{aligned}F\left[ f\right] =\int ^{\infty }_{-\infty }f\left( x\right) e^{ikx} dx\\ \Rightarrow F\left[ f\right] =A\int ^{\infty }_{-\infty }\cos xe^{ikx}dx\\ \Rightarrow F\left[ f\right] =A\int ^{\infty }_{-\infty }\cos x\left( \cos kx+i\sin kx\right) dx\\ \Rightarrow Re\left[ F\left( f\right) \right] =A\int ^{\infty }_{-\infty }\cos x\cos kxdx\end{aligned}
$\Rightarrow Re\left[ F\left( f\right) \right] =A\int ^{\infty }_{-\infty }\dfrac{1}{2}\left[ \cos \left( x+kx\right) +\cos \left( x-kx\right) \right] dx $
$\Rightarrow Re\left[ F\left( f\right) \right] =\dfrac{A}{2}\left[ \dfrac{\sin \left( 1+k\right) x}{1+k}+\dfrac{\sin \left( 1-k\right) x}{1-k}\right] _{-\infty }^{\infty } $
But $\sin(x)$ is not defined at $\infty$ and $-\infty$. So, how to find the Fourier transform?
EDIT: We can also do the following: \begin{aligned}F\left[ f\right] =A\int ^{\infty }_{-\infty }\cos xe^{ikx}dx\\ \Rightarrow F\left[ f\right] =\dfrac{A}{2}\int ^{\infty }_{-\infty }\left( e^{ix}+e^{-ix}\right) e^{ikx}dx\\ \Rightarrow F\left[ f\right] =\dfrac{A}{2}\int ^{\infty }_{-\infty }\left\{ e^{i\left( k+1\right) x}+e^{i\left( k-1\right) x}\right\} dx\\ \Rightarrow F\left[ f\right] =\dfrac{A}{2}\left[ \delta\left( k+1\right) +\delta \left( k-1\right) \right] \end{aligned} But how do we show that $\int ^{\infty }_{-\infty }e^{i\left( k\pm 1\right) x}dx=\delta \left( k\pm 1\right) $
It comes from the fact that:
$$\int_{-\infty}^{\infty} f(x) \delta(x-a) \, dx = f(a).$$
Then, if you have $F[f] = \frac{1}{2}\left(\delta(\omega + 1) + \delta(\omega - 1) \right)$ and plug it into the formula for the inverse Fourier transform, you get:
\begin{align} F^{-1}\lbrace F[f] \rbrace &= \int_{-\infty}^{\infty} \frac{1}{2}\left(\delta(\omega + 1) + \delta(\omega - 1) \right) e^{i\omega t} d\omega \\ &= \frac{1}{2} \left(e^{it} + e^{-it} \right) = \cos(t). \end{align}