How to find the second time when a piezoelectric crystal vibrates given a cubic trigonometric equation? The problem is as follows:

In an electronics factory a quartz crystal is analyzed to get its
vibration so this frequency can be used to adjust their components.
The length that it expands from a certain charge is given by:
$6\tan\left(\frac{\pi t}{24}\right)\cdot\tan\left(\frac{\pi
 t}{24}+\frac{\pi}{3}\right)\cdot\tan\left(\frac{\pi
 t}{24}-\frac{\pi}{3}\right)$ micrometers.
Assuming this interval $0\leq t \leq 12$. Where $t$ is measured in
seconds. Find on which second the crystal expands $6$ micrometers?

I'm not sure how to solve this question because I'm ending with a cubic equation and I don't really know how to solve that using precalculus tools.
The thing is that all that equation is:
$6\tan\left(\frac{\pi t}{24}\right)\cdot\tan\left(\frac{\pi t}{24}+\frac{\pi}{3}\right)\cdot\tan\left(\frac{\pi t}{24}-\frac{\pi}{3}\right)=6$
Hence:
$\tan\left(\frac{\pi t}{24}\right)\cdot\tan\left(\frac{\pi t}{24}+\frac{\pi}{3}\right)\cdot\tan\left(\frac{\pi t}{24}-\frac{\pi}{3}\right)=1$
$\tan\left(\frac{\pi t}{24}\right)\cdot\left[\frac{\tan\frac{\pi t}{24}+\sqrt{3}}{1-\sqrt{3}\tan\frac{\pi t}{24}}\right]\cdot\left[\frac{\tan\frac{\pi t}{24}-\sqrt{3}}{1+\sqrt{3}\tan\frac{\pi t}{24}}\right]=1$
Then:
$\tan\left(\frac{\pi t}{24}\right)\cdot\left[\frac{\tan \frac{\pi t}{24}-3}{1-3\tan^2\frac{\pi t}{24}}\right]=1$
$\tan\left(\frac{\pi t}{24}\right)\cdot\left[\tan \frac{\pi t}{24}-3\right]=1-3\tan^2\frac{\pi t}{24}$
$\tan^3 \left(\frac{\pi t}{24}\right)-3 \tan \left(\frac{\pi t}{24}\right)=1-3\tan^2\left(\frac{\pi t}{24}\right)$
$\tan^3 \left(\frac{\pi t}{24}\right) +3\tan^2\left(\frac{\pi t}{24}\right) -3 \tan \left(\frac{\pi t}{24}\right) -1 =0$
Then I landed here, now what?
Because I'm not familiar with solving this sort of equation it would help me alot someone could guide on me what should be done next.
The key from what I could notice is that if I could get that time, I would find when it gets its second value, hence the second time. But as I mentioned. I don't know how to do that.
 A: You made the same mistake a few times: you neglected the $t$ in $\tan\frac{\pi t}{24}$ in some of your equations, which may be the cause of your confusion. We have the equation
$$x^3+3x^2-3x-1=0$$ where $x=\tan\frac{\pi t}{24}$. Simply by inspection, after trying a few different values for $x$ we see that $x=1$ is a solution; hence by the factor theorem
$$\begin{align}
x^3+3x^2-3x-1&\equiv(x-1)(Ax^2+Bx+C)\\
&\equiv Ax^3+(B-A)x^2+(C-B)x-C
\end{align}$$
We are now in a position to compare coefficients. We can first see that $A=1$. Also, $B-A=3$, but since $A=1$ this means that $B=4$. Similarly, we can see that the constant terms on both sides must be equal so finally we have $C=1$, and thus
$$
x^3+3x^2-3x-1\equiv(x-1)(x^2+4x+1)$$
This is now something that we can solve, as we have two factors, one of which is simply linear and hence very easy to deal with, and the other quadratic which can be dealt with by employing the quadratic formula.

I think you must have made an error: the quadratic factor is factorisable over the reals, as shown below.
$$x^2+4x+1=(x-(-2+\sqrt 3))(x-(-2-\sqrt 3))$$
I got these values by using the quadratic formula- I didn't work them out just by looking at the equation :) These are actually very nice factors in our case, as... drumroll...
$$\tan\left(\frac{7\pi}{12}\right)=-2-\sqrt3~~~~\tan\left(\frac{11\pi}{12}\right)=-2+\sqrt3$$

I hope that was helpful. If you have any questions please don't hesitate to ask :)
