Considering the damped sinusoid $$x(t)=Ae^{-at} \cos{\omega t}\tag{1}$$
I would like to obtain the exact ranges of t where the second derivative is negative.
$$x'(t)=-aAe^{-at}\cos{\omega t} - Ae^{-at} \omega \sin {\omega t}\tag{2}$$
Note that this first derivative is zero when $\tan{\omega t}=\frac{-a}{\omega}\implies t=\frac{tan^{-1}{\frac{-a}{\omega}}+k\pi}{\omega},k\in \mathbb{Z}$
Note also that if we consider $t$ a function of $k$ we can calculate the difference in t between two successive critical points: $t(k+1)-t(k)=\frac{\pi}{\omega}$. Because we have a sinusoid, this is the distance in t between a max and a min or vice versa.
Now I calculate when the second derivative is zero
$$x''(t)=a^2Ae^{-at}\cos{\omega t}+aAe^{-at} \omega \sin{\omega t} +aAe^{-at} \omega \sin{\omega t} - Ae^{-at} \omega^2 \cos{\omega t}\tag{3}$$ $$x''(t)=Ae^{-at}(\cos{\omega t}(a^2-\omega^2)+2a\omega\sin(\omega t))\tag{4}$$ $$x''(t)<0 \implies \cos{\omega t}(a^2-\omega^2)+2a\omega\sin(\omega t) < 0\tag{5}$$ $$-\tan{\omega t}>\frac{a^2-\omega^2}{2a\omega}\tag{6}$$ $$\tan{\omega t}<\frac{\omega^2-a^2}{2a\omega}\tag{7}$$ $$t<\frac{tan^-1\big({\frac{\omega^2-a^2}{2a\omega}}\big)}{\omega}\tag{8}$$
I am not sure how to interpret this result.
If we take a specific case where $a=0.25$ and $\omega=1$ we have $x(t)=e^{-0.25t}$, we have:
The condition we derived for $x''(t)<0$ is $t<tan^-1{\frac{1-0.25^2}{2*0.25*1}}=1.08$.
From the graph and intuition, it seems that there is an inflection point every time the graph of $x(t)$ crosses $x=0$. From this point of view, the sign of the second derivative changes every $\frac{\pi}{\omega}$ length of t, or in our example every length $\pi$ in t. Here is a graph of $$\tan{\omega t}<\frac{\omega^2-a^2}{2a\omega}=\frac{1-0.25^2}{2*0.25*1}$$
Clearly, $tan(\omega t)$ spends most of its time below $\frac{\omega^2-a^2}{2a\omega}$.
Considering the result obtained above, namely $$t<\frac{tan^-1\big({\frac{\omega^2-a^2}{2a\omega}}\big)}{\omega}$$
How do we get this result that the sign of second derivative is half the time positive and half the time negative? This result of t smaller than a certain value seems to imply otherwise, namely that the second derivative spends much more time being negative than positive.
Let's start at equation $(5)$ from the original question:
$$x''(t)<0 \implies \cos(\omega t)(a^2-\omega^2)+2a\omega sin(\omega t)<0$$ $$\cos(\omega t)(a^2-\omega^2)=-2a\omega sin(\omega t)$$
Case 1) $\cos(\omega t)>0 \implies -\frac{\pi}{2} < \omega t <\frac{\pi}{2}$
$$\infty>-\tan(\omega t)>\frac{a^2-\omega^2}{2a\omega}$$ $$-\infty <\tan(\omega t)<\frac{\omega^2-a^2}{2a\omega}$$ $$-\frac{\pi}{2 \omega}<t<\frac{\tan^{-1}{\frac{\omega^2-a^2}{2a\omega}}}{\omega}$$
Case 2) $\cos(\omega t)<0 \implies \frac{\pi}{2} < \omega t <\frac{3\pi}{2}$
$$-\infty<-\tan(\omega t)<\frac{a^2-\omega^2}{2a\omega}$$ $$\infty >\tan(\omega t)>\frac{\omega^2-a^2}{2a\omega}$$ $$\frac{3\pi}{2 \omega}>t>\frac{\tan^{-1}{\frac{\omega^2-a^2}{2a\omega}}}{\omega}$$
We can visualize the two cases. The green portion below is the result of Case 1): it is range of $t$ in the interval $-\frac{\pi}{2\omega} < t <\frac{\pi}{2\omega}$ in which the second derivative is negative. The purple portion is the result of Case 2): the range of $t$ in the interval $\frac{\pi}{2\omega} < t <\frac{3\pi}{2\omega}$ in which the second derivative is negative. Note also that I added the red portion, which is where the second derivative is positive.
Indeed, the second derivative is negative half the time and positive half the time.
Another way to visualize this: