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.
 A: Since your pendulum oscillates about a new angle, call it $\theta_0$, your Taylor series approximation should be about $\theta_0$. So,
\begin{align*}
l \ddot{\theta} & =  -g \sin{\theta} + A \cos{\theta} \\
& \approx -g ( \sin{\theta_0} + (\theta - \theta_0) \cos{\theta_0} ) + A ( \cos{\theta_0} - (\theta - \theta_0) \sin{\theta_0}) \text{ for small deviations from } \theta_0 \\
& = A \cos{\theta_0} - g \sin{\theta_0} + \theta_0 (g \cos{\theta_0} + A \sin{\theta_0}) - \theta (g \cos{\theta_0} + A \sin{\theta_0})
\end{align*}
Solving that should give you the period you're looking for.
A: Personal choice, I would change the frame such that $\vec{g}' = \vec{g} - \vec{A}$. Doing this means gravity is larger and the frame is a little rotated, so choose a frame that lines up with gravity and say it's all the same. Then choose $\theta=0$ to be parallel to  $\vec{g'}$, you get the same equations as usual, where $g' = \big|\vec{g}'\big|$.
$$
\ddot{\theta}+\frac{g'}{l}sin\theta = 0
$$
A: I think the equations should be
\begin{cases}
    ml\ddot{\theta} = -mg\sin(\theta ) + mA\sin(\theta ) \\
    m\dot{\theta}^{2}l = T - mg\cos(\theta) - mA\cos(\theta)
\end{cases}
due to 
\begin{equation}
\vec{g}=-g\cos(\theta )\hat{\rho} - g\sin(\theta )\hat{\theta} 
\end{equation}
 and 
\begin{equation} 
\vec{A} = A\hat{x} = A\cos(\theta )\hat{\rho} - A\sin(\theta )\hat{\theta}
\end{equation}
