While working on some Rolle's Theorem problems I came to:
$$f(x) = \cos 3x$$
This is both continuous on the given interval (and everywhere really) $[{\pi\over 12}, {7\pi \over12}]$ and differentiable $({\pi\over 12}, {7\pi \over12})$. Additionally, $f({7\pi \over12}) = f({\pi\over 12})$. So all three hypotheses check out.
So now I need to find the c-values on the open interval. So I take the derivative and solve for zero:
$$0 = -\sin 3x \cdot3$$
However, my trig is very rusty. How do I solve this? I figured out the answer as ${4\pi\over 12}$ by determining what I would need to get $sin\space \pi$ (i.e. $\pi= 3\cdot({\pi\over12})$). Is this how you solve this type of trig problem, or is there a better, more formal, robust, standard way?
First of all, you can obviously divide through by $-3$ to get the equation $\sin(3x)=0$.
From here it's easy: $\sin(y)=0$ if and only if $y=k\pi$ for some integer $k$. So if we let $y=3x\ldots$
Once we have all the solutions like so, it becomes obvious that the only solution in your chosen interval is $x=4\pi/12$