How do I derive the angular frequency of a simple pendulum through conservation of energy? Is it possible? I'm not exactly sure what I'm doing wrong. 
So far I've gotten:
$mgl(1-$cos$\theta) = \frac12\omega^2l^2$
Which then gives $\omega = \sqrt\frac{2g(1-cos\theta)}{l}$ which is incorrect. Where am I going wrong??
 A: The problem is $v=wr$ only works for constant acceleration case, for rotational motion. So we cannot just say $d\theta /dt=\omega$.
The total energy of the system 
$$E=1/2m\dot{\theta}^2 l^2+mglcos(\theta)$$
Since theres no external force acting on the system.
$dE/dt=0$
Hence
$$m\dot{\theta} l^2\ddot{ \theta}+mglsin(\theta)\dot{ \theta}=0$$ 
Since $v=\dot{ \theta}l \neq 0$ 
We can write,
$$m\dot{\theta} l[\ddot{ \theta}l+g\theta]=0$$ 
And by setting $sin(\theta)=\theta$ using small angle approximation.
Hence 
$$\ddot{ \theta}l+g\theta=0$$
Or $$\ddot{ \theta}+\omega^2 \theta=0$$
For $\omega=\sqrt{g/l}$
A: In your equation, $\omega$ stands for angular speed, not angular frequency. The expression you found gives you the angular speed at the bottom of the swing as a function of the maximum angle to vertical $\theta$.
Note: you forgot an $m$ on the right hand side of your original equation, but I figure that was a typo since you cancelled it out correctly in your result.
Unfortunately, your equation simply equates initial and final energies, and provides no information about anything in between the initial and final states. Thus, the equation does not contain sufficient information to determine the angular frequency.

What could, however be used to determine the frequency, would be an equation that tells you the kinetic energy $K$ as a function of the angle $\theta(t)$ ,which in turn is a function of time (and thus provides you information about all the points in time between the initial and final states). Notice that I'm now using $\theta$ now in a different way than you were in your equation. To distinguish, let's call the maximum angle $\theta_\text{max}$. 
$$K(\theta) = mgl(1 - \cos \theta_\text{max}) - mgl(1 - \cos \theta) = mgl(\cos \theta - \cos \theta_\text{max})$$
Set this equal to $K = \frac{1}{2} m \omega^2 l^2 = \frac{ml^2}{2} \left( \frac{d\theta}{dt} \right)^2$.
$$mgl(\cos \theta - \cos \theta_\text{max}) = \frac{ml^2}{2} \left( \frac{d\theta}{dt} \right)^2$$
$$\left( \cos \theta - \cos \theta_\text{max} \right) = \frac{l}{2g} \left( \frac{d\theta}{dt} \right)^2$$
Assume a small angle oscillation $\theta_\text{max} \approx 0$ and substitute the small angle approximations $\cos \theta_\text{max} \approx 1 - \theta_\text{max}^2$ and $\cos \theta \approx 1 - \theta^2/2$.
$$\theta_\text{max}^2 - \theta^2 = \frac{l}{g} \left( \frac{d\theta}{dt} \right)^2$$
$$\frac{l}{g}\left( \frac{d\theta}{dt} \right)^2 + \theta^2 = \theta_\text{max}^2$$
This relationship looks reminiscent of the Pythagorean Theorem. Note that the two terms on the left hand side are time dependent, whereas the terms on the right hand side are time independent. These two hints suggest that the solution $\theta(t)$ will involve a trig function (since it's an oscillation, this seems exceedingly sensible). So, let's input he trial solution $\theta(t) = \theta_\text{max} \sin \omega t$, where $\omega$ is the angular frequency.
$$\frac{l}{g}\left( \theta_\text{max} \omega \cos \omega t \right)^2 + (\theta_\text{max} \sin \omega t)^2 = \theta_\text{max}^2$$
$$\frac{l\omega^2}{g}\left( \cos \omega t \right)^2 = 1 - (\sin \omega t)^2$$
$$\frac{l\omega^2}{g}\left( \cos \omega t \right)^2 = (\cos \omega t)^2$$
$$\frac{l\omega^2}{g} = 1$$
$$\omega^2 = \frac{g}{l}$$
$$\boxed{\omega = \sqrt{\frac{g}{l}}}$$
