Does sinc function have any special inverse function defined? We know that $y=xe^x$ cannot be solved for $x$ using elementary functions.
The Lagrange inversion theorem can be used for finding a "new" function that would be the inverse function of the above equation. This special function is named "Lambert W Function"
So for $y=xe^x$, $x=W(y)$.
There are many equations that can be solved through Lambert's W Function.  However it seems that some common equations in Optics or Control Theory, like $y=\dfrac{\sin x}{x}$  or $y=e^{-x}\cos x$ cannot be solved with Lambert W Function. 
I wonder if there are already any specials, Lambert-W-like, functions for those cases, or their inverse functions still remain undefined.
 A: For a function to be invertible it must be monotonic.  y = $xe^x$ is monotonic.  However, sinx/x and $e^{-x}$cosx are monotonic only in small intervals.  So you certainly can't have a universal inverse for either of them.  
A: There is an experimental series expansion using the Lagrange Reversion theorem:

$$x=a+bf(x)\implies x=a+\sum_{n=1}^\infty\frac{b^n}{n!}\frac{d^{n-1}}{da^{n-1}}(f^n(a))$$

If we take $f(x)=\cos(x),a=0$, then we get:
$$x=b\cos(x)\implies \frac1b=\frac{\cos(x)}x$$
and use the formula to get the inverse of a similar function since choosing $f(x)=\sin(x)$ gives series coefficients all as $0$:
$$x=\sum_{n=1}^\infty\frac{b^n}{n!}\left.\left(\frac{d^{n-1}}{da^{n-1}}(\cos^n(a))\right)\right|_{a=0}$$
The formula works unexpectedly
Now use the binomial theorem and the complex definition of cosine:
$$\sum_{n=1}^\infty\frac{b^n}{n!}\left.\left(\frac{d^{n-1}}{da^{n-1}}(\cos^n(a))\right)\right|_{a=0} = \sum_{n=1}^\infty\frac{b^n}{n!}\left.\left(\frac{d^{n-1}}{da^{n-1}}\left(2^{-n}\sum_{k=0}^n\binom nke^{(2ik-in)a}\right)\right)\right|_{a=0}$$
There are derivative patterns:
$$\sum_{n=1}^\infty\frac{b^n}{2^nn!} \sum_{k=0}^n\binom nk (2ik-in)^{n-1}$$
If these $n$th derivatives are correct, then we have a working formula for an inverse of $\frac {\cos(x)}x$ by substituting $\frac1b\to b$:
$$\boxed{\frac{\cos(x)}x=b\implies x=\sum_{n=1}^\infty\sum_{k=0}^n\frac{(2ik-in)^{n-1}}{k!(n-k)!(2b)^n}}$$
Which works possibly for $|b|>b_0$ where $b_0\approx1.8$
Here is an approximated complex plot of an inverse $\frac{\cos(z)}z$ function:

Wolfram Alpha fails to find the coefficients as a function and the Kepler equation series solution fails too since the coefficients are all $0$ using it. However Wolfram Alpha finds this pattern with the Hypergeometric function
$$\sum_{k=0}^n\binom nk (c+g k)^m=c^m\,_{m+1}\text F_m\left(\underbrace{\frac cg+1,…,\frac cg+1}_{m\text{ times}},-n;\underbrace{\frac cg,…,\frac cg}_{m\text{ times}};-1\right),m\in\Bbb N$$
now we set $m=n-1,c=2i,g=-in$ to get a possible solution since all terms are defined:
$$\begin{align}\sum_{n=1}^\infty\frac1{(2b)^nn!} \sum_{k=0}^n\binom nk (2ik-in)^{n-1} = \sum_{n=1}^\infty\frac1{(2b)^nn!}  (2i)^{n-1}\,_n\text F_{n-1}\left({1-\frac 2n,…,1-\frac 2n},-n;{-\frac 2n,…,-\frac2n};-1\right)\iff \frac{\cos(x)}x=b\implies x=\frac 12\sum_{n=1}^\infty\frac{i^{n-1}}{b^n n!}\,_n\text F_{n-1}\left({1-\frac 2n,…,1-\frac 2n},-n;{-\frac 2n,…,-\frac2n};-1\right)\end{align}$$
Using the Kepler equation series solution, we simplify the hypergeometric function into Bessel J:
$$\boxed{\frac{\cos(x)}x=b\implies x=-2\sum_{n=1}^\infty \text J_n\left(-\frac n b\right)\frac{\sin\left(\frac\pi 2n\right)}n}$$
Unfortunately, using the Bessel J integral representation gives a trivial floor function integral representation for the inverse function with the integrand in the link
A: My intuition for the lack of inverses for your two functions is that they have infinite branches, and the infinite branches are not in a simple relation. Contrast with Lambert W function which has two branches, and $\sin^{-1}$ which the branches are related in simple enough periodic manner.
