In my original comment, I suggested solving your problem using Lagrangian mechanics. However, I realized you can do this using forces if you change into polar coordinates. Newton's second law, in polar coordinates, has the form
\begin{equation}
\vec F_{net} = m(\ddot r - r\dot\phi^2)\hat r + m(2\dot r \dot \phi + r\ddot \phi)\hat \phi
\end{equation}
where $\hat r$ is unit vector that points along the pendulum's rod and $\hat \phi$ is a unit vector parallel to the bob's velocity vector. By drawing a force diagram, we find that
$$ \vec F_{net} = (F_T - mg\cos\phi)\hat r + (-mg\sin\phi)\hat \phi,$$
where $F_T$ is the force from the rod. Except that for your purposes, we don't really care about the $\hat r$ direction too much. Since our unit vectors are linearly independent, we can ignore the $\hat r$ direction and combine these two equations to find
$$ -mg\sin\phi = m(2\dot r\dot\phi + r\ddot\phi).$$
Knowing that $\dot r$ is zero (the rod length doesn't change), we can rewrite this to find
$$ \ddot\phi = - \frac{g}{r}\sin\phi, $$
where in your notation $l = r$ and $\phi = \theta$. Note that this is exactly what you find by solving the Euler-Lagrange equations. As it stands, this DE cannot be solved analytically. However, we can constrain the pendulum to small angles such that
$$ \ddot \phi = - \frac{g}{r}\phi, $$
a linear, homogeneous differential equation which we can solve to find
$$ \phi(t) = c_1\sin(\sqrt{\frac{g}{r}}t) + c_2\cos(\sqrt{\frac{g}{r}}t),$$
where $\sqrt\frac{g}{r}$ is the angular frequency of the pendulum. You may notice that torque and angular momentum are somewhat implicit in this derivation. However, we solved the problem using only a force diagram, which is what I perceive to be your question.
