homework and exercises - Period of simple pendulum accelerated horizontally

I'm confused about simple pendulum problems where the pendulum is accelerated horizontally of anyway not vertically with acceleration $\vec{A}$.

$m\vec{g} + \vec{T}-m \vec{A} =m \vec{a}$

So

$\begin{cases} m l \ddot{\theta} = - mg sin(\theta)+m A cos(\theta) \\ m \dot{\theta} ^2 l = T - mg cos(\theta)-m A sin(\theta) \end{cases}$

From the first equation, on the tangential coordinate,

$l \ddot{\theta} = - g sin(\theta)+ A cos(\theta)$

Which is for small angles

$l \ddot{\theta} = - g \theta+ A$

And therefore the period of small oscillations should still be $\tau=\sqrt{\frac{l}{g}} 2\pi$

While of course it is different, but I don't see the mistake in what I wrote here.