Find how many solutions are to $(x-3)(x-2)=\sin x$ 
Find how many solutions exist to the equation $(x-3)(x-2)=\sin x$

Let $E(x) = (x-3)(x-2)-\sin x$.
$E(0)>0,\quad E(1,5) < 0, \quad E(3) > 0$
Because $E(x)$ is continuous and changes signs 3 times according to Intermediate Value theorem there're at least 2 roots. 
$E(x)$ is differentiable for all $x$ if there're 3 or more roots by Rolle's theorem $E'(x)$ would have 2 roots.
$E'(x) = 2x-5-\cos x$
$2x = 5+\cos x$
How do I prove that $2x = 5+\cos x$ has less than 2 roots? Or did I do something wrong along the way?

The analogy that you are using only applies to polynomials.
For example, let the function $f(x) = 5x^4 + 3$ 
and let $g(x)$ be a primitive of $f(x)$. Then we have that $g(x) = x^5 + 3x + C$. 
Depending on the constant $C$, the number of roots will change. 
Likewise, let $f(x) = \cos{x} + 10$. This equation does not have any roots, however, the preceding derivatives have an infinite number of roots.
