I would like to find the real roots of the function $$i(t) = \frac{\hat{V}}{R}\left(\frac{\omega^2}{(\alpha^2 + \omega^2)} \cos\left(\omega t + \tan^{-1}\left(\frac{\alpha}{\omega}\right)\right) + \frac{\alpha^2}{(\alpha + \omega)^2} e^{-\alpha t}\right)$$
in the form $t = f\left(\hat{V}, R, \omega, \alpha\right)$, where $\hat{V}, R, \omega$ and $\alpha$ are real constants.
Setting $i(t)=0$ and simplifying has brought me to the equation
$$-\frac{\alpha^2}{\omega^2} = e^{\alpha t}\cos(\omega t + \phi)$$
where $\phi = \tan^{-1}\left(\frac{\alpha}{\omega}\right)$.
How can this be solved for $t$?
As I said in a comment, I do not think that you could obtain an explicit solution for $$f(t)= e^{\alpha t}\cos(\omega t + \phi)+\frac{\alpha^2}{\omega^2}=0$$ where $\phi = \tan^{-1}\left(\frac{\alpha}{\omega}\right)$.
However, if you look for the first solution, you could expand $f(t)$ as a Taylor series around $t=0$ and obtain $$f(t)=\left(\frac{\alpha ^2}{\omega ^2}+\frac{1}{\sqrt{\frac{\alpha ^2}{\omega ^2}+1}}\right)-\frac{t^2 \left(\alpha ^2+\omega ^2\right)}{2 \sqrt{\frac{\alpha ^2+\omega ^2}{\omega ^2}}}+O\left(t^3\right)$$ So, a first approximation of the first solution could be given by $$t^2=\frac{2 \left(k^2 \sqrt{k^2+1} +1\right)}{\left(k^2+1\right) \omega ^2}$$ where $k=\frac{\alpha}{\omega}$.
Continuing the expansion for an extra term would lead to a quite unpleasant cubic equation in $t$.
What is also interesting to notice is that, after simplification, we can write $$f'(t)=-\sqrt{\alpha ^2+\omega ^2} \sin (t \omega ) e^{\alpha t}$$ which allows to see the location of all maximum and minimum of $f(t)$ which just correspond to $ \sin (t \omega )=0$ so $\omega t=n \pi$.