Need help with proof on trigonometric integrals Question:
Let $f:[0,\pi]\rightarrow\mathbb{R}$  be a continuous function. Show that, if $$\int\limits_0^{\pi} f(t)\sin(t)dt = \int\limits_0^{\pi} f(t)\cos(t)dt =0,$$ then the equation $f(x)=0$ admits at least two solutions in $[0,\pi].$
My thoughts:
If $f(x) > 0$, then $f(x)\sin(x) > 0 \hspace{1em}\forall x \in(0, \pi)$,
but $f(x)\cos(x) > 0 \hspace{1em}\forall x \in[0,\frac{\pi}{2})\hspace{1em}$ and $\hspace{1em}f(x)\cos(x) < 0 \hspace{1em}\forall x \in(\frac{\pi}{2},\pi]$
I'm not really sure how to proceed.
 A: As observed in the comments, $f$ must have at least one zero: otherwise $f$ has no zeroes, and by continuity $f$ is always positive, or always negative. Suppose that $f>0$. Recalling that $\sin x \ge 0$ for $x \in [0, \pi ]$ you get
$$\int_0^\pi f(x) \sin x \ \mathrm dx >0$$
Now, suppose by contradiction that $f$ has exactly one zero. Such a zero must be in the interior of $(0, \pi)$, otherwise we would have a contradiction using the same argument above.
Say $f( A)=0$ is the unique zero of $f$, with $A \in (0, \pi)$. Since $f$ is continuous, it changes sign only once. Without loss of generality we can suppose that $f >0$ in $[0, A)$, and $f<0$ in $(A, \pi]$.
Recall that for all $x \in \Bbb R$ you have
$$\sin (x-A) = \sin x \cos A - \cos x \sin A$$
In particular, for all $x \in [0, \pi]$ you can multiply by $f(x)$, and the you can integrate over $[0, \pi ]$ to get
$$\int_0^\pi f(x) \sin (x-A)  \ \mathrm dx = 0$$
However this cannot happen. Indeed for all $x \in [0, A)$ you have
$$\sin (x-A) < 0$$
while for all $x \in (A, \pi]$ you have
$$\sin (x-A) > 0$$
This means that for all $x \neq A$ you have $f(x) \sin (x-A) <0$. Hence
$$0=\int_0^\pi f(x) \sin (x-A)  \ \mathrm dx <0$$
A contradiction.
