We would like to rewrite $\cos 2t +\sqrt{3}\sin 2t$ in this form
$$\cos\alpha\cos 2t-\sin\alpha\sin2t$$
that way, we could simplify the expression to $\cos(\alpha + 2t)$.
Following this model, we would have $\cos\alpha=1$ and $\sin\alpha=-\sqrt{3}$. At this point, we find a contradiction in our argument because
$$1 = \cos^2\alpha + \sin^2\alpha = 1^2+(-\sqrt{3})^2=4$$
To fix the problem, we need to scale by a factor of $\sqrt{4}$, so lets define $\alpha$ as $\cos\alpha = \frac{1}{2}$ and $\sin\alpha=-\frac{\sqrt{3}}{2}$.
Now we have
$$\cos^2\alpha + \sin^2\alpha = (\frac{1}{2})^2+(-\frac{\sqrt{3}}{2})^2 = \frac{1+3}{4}=1$$
so we have that
$$\cos\alpha\cos 2t-\sin\alpha\sin2t = \frac{1}{2}\cos 2t +\frac{\sqrt{3}}{2}\sin 2t$$
scaling this expression by a factor of $2$, or $\sqrt{1^2+\sqrt{3}^2}$, we have our original expression back
$$2(\frac{1}{2}\cos 2t +\frac{\sqrt{3}}{2}\sin 2t)=\cos 2t +\sqrt{3}\sin{2t}$$
Now we find alpha, and plug it into our expression (notice the acrobatics with the signs)
$$\begin{array}{lll}
2(\frac{1}{2}\cos 2t +\frac{\sqrt{3}}{2}\sin 2t)&=&2(\cos(-\frac{\pi}{3})\cos 2t - \sin(-\frac{\pi}{3})\sin 2t)\\
&=& 2\cos(2t -\frac{\pi}{3})\\
&=&2\cos(2(t - \frac{\pi}{6}))
\end{array}$$