“The Fourier transform cannot measure two phases at the same frequency.” Why not?
It's because the simultaneous presence of two sinusoidal signals with the same frequency and different phases is actualy equivalent to  a single sinusoidal at the same frequency, but, with a new phase and amplitude as follows:
Let the two sinusodial components be summed like this :
$$ x(t) = a \cos(\omega_0 t + \phi) + b \cos(\omega_0 t + \theta) $$
Then, by trigonometric manipulations it can be shown that :
$$ x(t) = A \cos(\omega_0 t + \Phi) $$
where
$$A = \sqrt{ a^2 + b^2 + 2 a b \cos(\theta-\phi) } $$ and
$$ \Phi = \tan^{-1}\left(\frac{ a \sin(\phi) + b\sin(\theta) } { a \cos(\phi) + b\cos(\theta) } \right) $$
hence you actually have a single sinusoidal (with a new phase and amplitude), and therefore nothing to distinguish indeed...