I don't know how to attack this problem. The last I've tried is using a differential equation, but I don't know how to solve it.
Let $y$ be $f^{-1}(x)$. Knowing that $x=f(y)= 2y + \cos(y)$ and derivating I obtained the following non-linear first order differential equation: $y' \cdot (2-\sin(y))=1$
I would thank you if you can help me.
Edit: I haven't said, but it is trivial to check that the function is injective, so it has an inverse, because $\forall x \in \mathbb{R}$ $f'(x) \neq0$
Best Answer
Given function $$ y= 2 x + \cos x \tag{1} $$ Inverse function $$ x = 2y + \cos y \tag{2}$$ Differentiating with respect to $x$ $$ 1 = 2 y^{'} - \sin y \, y^{'} $$ $$ y^{'} = \dfrac{1}{2-\sin y} $$
Cannot be further put in terms of $x$ as (1) and (2) are transcendental.