The Mean Value Theorem states:
If $f(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists at least on $c\in (a,b)$ such that
$$f'(c) = \frac{f(b)-f(a)}{2\pi}.$$
Our function here is $f(x) = x+\sin(2x)$, and $[a,b]$ is $[0,2\pi]$.
The function is continuous everywhere: $x$ is continuous everywhere, so $2x$ is continuous everywhere. Since $\sin(u)$ is continuous everywhere, the composition $\sin(2x)$ is continuous everywhere. Since $x$ and $\sin(2x)$ are both continuous everywhere, their sum is continuous everywhere. And since $f(x) = x+\sin(2x)$ is continuous everywhere, it is in particular continuous on $[0,2\pi]$.
Similarly, each of the functions mentioned is differentiable everywhere, so $f(x)=x+\sin(2x)$ is differentiable everywhere. Since it is differentiable everywhere, it is also differentiable on $(0,2\pi)$.
According to the Mean Value Theorem, there must exist at least one point $c$ in $(0,2\pi)$ where
$$\begin{align*}
f'(c) &= \frac{f(2\pi)-f(0)}{2\pi - 0}\\
&= \frac{\Bigl( 2\pi + \sin(2(2\pi))\Bigr) - \Bigl( 0 + \sin(2(0))\Bigr)}{2\pi}\\
&=\frac{2\pi + \sin(4\pi) - 0 - \sin (0)}{2\pi}\\
&= \frac{2\pi + 0 - 0 + 0}{2\pi}\\
&=\frac{2\pi}{2\pi}\\
&= 1.
\end{align*}$$
So the Mean Value Theorem tells us that there is at least one point $c$ in $(0,2\pi)$ where $f'(c) = 1$.
That would be it.
You may also want to verify that the conclusion is indeed true by exhibiting a point $c$ in $(0,2\pi)$ where this is true. We have
$$f'(x) = 1 + \cos(2x)(2x)' = 1 + 2\cos(2x).$$
So $f'(c) = 1$ if and only if $1+2\cos(2c) = 1$, if and only if $2\cos(2c)=0$, if and only if $\cos(2c)=0$.
Cosine is $0$ on the odd multiples of $\frac{\pi}{2}$; the values of $c$ on $(0,2\pi)$ where $\cos(2c)=0$ are $c=\frac{\pi}{4}$, $\frac{3\pi}{4}$, $\frac{5\pi}{4}$, and $\frac{7\pi}{4}$. So indeed, there is at least one (in fact four) points $c$ in $(0,2\pi)$ where $f'(c) = \frac{f(b)-f(a)}{b-a}$, as predicted by the Mean Value Theorem.
Polynomials are always going to satisfy theorem.
No. They satisy the hypothesis about continuity and differentiability in the closed and open intervals respectively while you need to check if the polynomials agree at the end point.
Here they will agree on the end points because $x^2$ and $x^4$ are even functions and the interval is symmetric about $0$. (Do you see why?)
And, you are in need of $c$ that satisfies $f'(c)=0$. You're right but I would be surprised if the instructor does not want you to prove your claim:
\begin{align}
f'(c)&=0 \;\;\mbox{and $c \in (-3,3)$} \tag{1}\\4c^3+8c&=0\\4c(c^2+2)&=0
\end{align}
Since $c^2+2 \ge 2$ for $c \in \Bbb R$, the equation $c^2+2=0$ does not have roots in $\Bbb R$ and hence $c=0$ is the only value that satisfies $(1)$.
Best Answer
We have to verify (copied from wikipedia...)
In this case, $f$ is continuous and differentiable, as it is the sum of terms that contain so-called $C^{\infty}$ functions.
$a = 0; b = 2 \pi$
$f(a) = f(0) = 0$;
$f(b) = f(2 \pi) = 2 \pi$.
Can you take the derivative and set it equal to $\dfrac{2 \pi - 0}{2 \pi - 0} = 1$ ?