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 $(0,\pi)$

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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.