Find acceleration when v(t) = 0 I am struggling with this...
This was a question I got wrong on a test, and I obviously did not even know how to solve it, so any help is greatly appreciated!
This is a simple velocity/acceleration question
$$s(t)=3\sin(2t-4)$$
Find v(t)
$$v(t)=6\cos(2t-4)$$
Find a(t)
$$a(t)=-12\sin(2t-4)$$
Find the acceleration of the particle when v(t)=0
Now, from what I understand, I need to set the velocity function equal to zero and solve for t, and then take the t value and plug it into my acceleration function... however, I am running into issues and getting very sloppy results. I am assuming I am doing something wrong, or am simply approaching it the wrong way.
$$6\cos(2t-4)=0$$
Where do I begin/How do I solve this?
Even when I consult WolframAlpha, I am getting the following answer for t
$$t=\frac{1}{4}(\pi(3-2n)+8)$$

To give the answer with the logic outlined in the comments:
$v(t) = 0$ implies $\cos(2t - 4) = 0$. Thus
$$2t - 4 = \pi/2 + k\pi \quad \text{ for some integer } k$$
In that case
$$a(t) = -12\sin(2t - 4) = -12\sin(\pi/2 + k\pi) = -12\cdot(-1)^k$$
Therefore
$$v(t) = 0 \ \Longrightarrow \ a(t) = \pm 12$$
