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
$$t=\frac{1}{4}(\pi(3-2n)+8)$$
 A: 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
$$v(t) = 0 \ \Longrightarrow \ a(t) = \pm 12$$
A: Let's begin with a fact: $2t-4$ is ugly. There is too much ugliness in the world, so let's instead say that $2t-4 = \theta$. Now this is better, we can instead write our equations as
$$\begin{align*}s(\theta(t)) &= 3\sin \theta(t), \\
v(\theta(t)) &= 3\cos \theta(t) \frac{d\theta}{dt}, \\
a(\theta(t)) &= -3\sin\theta(t)\left(\frac{d\theta}{dt}\right)^2 + 3\cos\theta(t) \frac{d^2 \theta}{dt^2}.
\end{align*}$$
Where'd these $\frac{d\theta}{dt}$ come from? We mustn't forget the chain rule! Fortunately for us, we have a simple case: $\frac{d\theta}{dt} = 2$ and $\frac{d^2\theta}{dt^2} = 0$.
This gives us:
$$\begin{align*}s(\theta(t)) &= 3\sin \theta(t), \\
v(\theta(t)) &= 3\cos \theta(t) \cdot 2, \\
a(\theta(t)) &= -3\sin\theta(t)\cdot 2^2 + 3\cos\theta(t) \cdot 0,
\end{align*}$$
which simplifies to
$$\begin{align*}s(\theta(t)) &= 3\sin \theta(t), \\
v(\theta(t)) &= 6\cos \theta(t), \\
a(\theta(t)) &= -12\sin\theta(t).
\end{align*}$$
Ah, now we're talking. This should be much easier. Suppose now that $v(\theta(t)) = 0$. As one knows, $\cos \theta = 0$ when $\theta = \frac{(2k+1)\pi}{2}$. But actually, this is irrelevant!! Why? Because when $\cos \theta = 0$, we have $\sin\theta = \pm 1$.
Therefore, from our relations as derived above, we must have that $a(t) = \pm 12$ whenever $v(t) = 0$.
