$\sin x$ does not satisfy this quadratic equation 
Prove that $\sin x$ is not a rational function using the fact that it is not of the form $p(x)/q(x)$ where $p$ and $q$ are polynomials. Then, by using the above proof, prove that $\sin x$ does not satisfy a "quadratic equation" of the form:
  $$
(\sin x)^2 f_2(x) + (\sin x)f_1(x) + f_0(x) = 0,
$$
  where $f_0, f_1, f_2$ are rational functions.

I know that a rational function cannot be zero at infinitely many points unless it is $0$ everywhere, but how does one use this information to formulate $p(x)/q(x)$ argument? If anybody could please help.
 A: For the second part, by multiplying the equation with a suitable polynomial, we may assume that $f_0$, $f_1$, $f_2$ are polynomials. We have:
$$
\forall n \in \Bbb Z : \sin^2(\pi n) f_2(\pi n) + \sin(\pi n) f_1(\pi n) + f_0(\pi n) = 0
$$
Therefore:
$$
f_0(\pi n) = 0
$$
Since a non-constant polynomial function can only have a finite number of zeros, $f_0$ must be $0$ everywhere. The equation reduces to:
$$
\sin^2(x) f_2(x) + \sin(x) f_1(x) = 0
$$
Assume $x \ne \pi n$ and factor out $\sin(x)$ to get:
$$
\forall x \ne \pi n : \sin(x) f_2(x) + f_1(x) = 0
$$
Which means that:
$$
\sin(x) = -\frac{f_1(x)}{f_2(x)}
$$
But $-f_1/f_2$ is also a rational function. By the continuity of $\sin$ and $-f_1/f_2$, equality must hold everywhere, which contradicts the fact that $\sin$ is not a rational function.
A: In fact, the curve $y = \sin x$ is not equal to (or contained in) any algebraic curve $f(x,y) = 0$, where $f(x,y)$ is any polynomial in two variables with real coefficients.  Indeed, the algebraic curve $f(x,y) = 0$ meets the $x$-axis in only finitely many points, but $y = \sin x$ meets the $x$-axis in infinitely many points.
