Prove that $ f(x) $ has at least two real roots in $ (0,\pi) $

Let $ f $ be a continuous function defined on $ [0,\pi] $. Suppose that

$$ \int_{0}^{\pi}f(x)\sin {x} dx=0, \int_{0}^{\pi}f(x)\cos {x} dx=0 $$

Prove that $ f(x) $ has at least two real roots in $ (0,\pi) $

Solutions Collecting From Web of "Prove that $ f(x) $ has at least two real roots in $ (0,\pi) $"

Here is one root:
Let $F(x) = \int_{0}^x f(t) \sin t dt$. Then $F(0)=0$ and $F(\pi)=\int_{0}^\pi f(t)\sin tdt=0$. So by the intermediate value theorem, there exists $0<c<\pi$ such that
$$
0=F'(c) = f(c)\sin c.
$$
But since $\sin c\neq 0$, we get that $f(c)=0$.

If f has only one real root on $ (0,\pi)$, say $ a \in (0,\pi) $, then define
$ g(x) = f(x) \sin(x-a) = f(x) (\sin(x)\cos(a) – \cos(x)\sin(a))$, then $ g(x) $ is either non-positive or non-negative, not identically zero, and has integral $ 0 $. Contradiction.