How to find the solutions to $\frac{11}{2} x - \cos x = 0$? 
Find the values of $x$ such that $\frac{11}{2} x - \cos x = 0$.

I really don't know how to find the solutions of this equation.  I would appreciate if you could help me.
 A: We can approximate the value of $x$ without Newton's Method or any calculus at all. We can generalize the idea behind solving for $x$ in $x = \cos(x)$ by repeatedly hitting the cosine button on a calculator and noticing that it converges.

Rearranging the equation a bit, we are looking for the $x$ such that 
$$
  x = \frac{2}{11}\cos(x)\;.
$$ 
Intuitively we know that such a value of $x$ exists since the graph of $y=\cos(x)$ and the line $y=\frac{11}{2}x$ intersect. 
Now let's consider the function $f$ defined as $f(\_) = \frac{2}{11}\cos(\_)$. Suppose that we repeatedly apply $f$ to some input $a$, and let this infinite composition equal $x$. So we are going to say
$$\begin{align}
   &f\left(f\left(\dotsb f(a)\right)\dotsb\right) 
   \\=\;\;&
   \frac{2}{11}\cos\left(\frac{2}{11}\cos\left(\dotsb \frac{2}{11}\cos(a)\right)\dotsb\right) 
   \\=\;\;& x
\end{align}$$
But if we suppose this converges as the number of times we apply $f$ approaches infinity (so $x$ actually exists and is a number), applying $f$ one more time shouldn't change the value of anything. So we now have 
$$\begin{align}
   &x
   \\=\;\;&
   f\left(f\left(\dotsb f(a)\right)\dotsb\right)
   \\=\;\;&
   f\left(f\left(f\left(\dotsb f(a)\right)\dotsb\right)\right)
   \\=\;\;&
   \frac{2}{11}\cos\left(\frac{2}{11}\cos\left(\frac{2}{11}\cos\left(\dotsb \frac{2}{11}\cos(a)\right)\dotsb\right)\right)
   \\=\;\;& \frac{2}{11}\cos(x)
\end{align}$$
So the $x$ that this may converge to, the $x$ that we are looking for, is precisely the $x$ the solves the equation $x = \frac{2}{11}\cos(x)$. Furthermore we know such an $x$ exists by the reasoning of intersecting graphs above! And since there is no $a$ in the equation $x = \frac{2}{11}\cos(x)$, that must mean the value of $x$ doesn't depend on $a$. This means we can approximate the value of $x$ in $x = \frac{2}{11}\cos(x)$ by taking any number and repeatedly applying $f$ to it.
We can now easily calculate this value of $x$ by programming the function $f(\_) = \frac{2}{11}\cos(\_)$ into a calculator repeatedly apply $f$ to some (any) input. Calculating $f^n$ for different values of $n$ (mostly to compare to Newton's Method in Claude Leibovici's answer) I get:
\begin{array}{c|c}
  n      & f^n(1) \\\hline
  1      & 0.\color{#A00}{0982367828} \\
  2      & 0.1\color{#A00}{809415720} \\
  3      & 0.178\color{#A00}{8499431} \\
  4      & 0.17891\color{#A00}{79884} \\
  5      & 0.178915\color{#A00}{7870} \\
  6      & 0.17891585\color{#A00}{83} \\
  \vdots & \\
  100    & 0.1789158560
\end{array}
So this doesn't converge as quickly as Newtons' Method, but I think that this can be punched into a calculator much quicker.
