Proving ω = √(g/L) for a pendulum

[Hootenanny]

I have already posted in the physics forum but its gone a bit quiet. I've managed to derrive an equation up to this point. I have $$\frac{g}{L}\theta = \omega^2 \theta_{max} \sin (\omega t - \alpha)$$ and I need to prove that $$\omega = \sqrt{\frac{g}{L}}$$. I'm stuped at this one. Thank's in advance for your help.

 

[moose]

well, you need to know that it's only an approximation. It is most accurate when sinx=x (in radians of course).

 

[Hootenanny]

That doesn't help me much all that gives then is $$\frac{g}{L}\theta = \omega^2 \theta_{max}(\omega t - \alpha)$$

 

[Hootenanny]

Now I have $$\theta = \theta_{max} \sin(\sqrt{\frac{g}{L}} t - \alpha )$$ How Can I remove the $\alpha$ ?