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Let $f\colon [0,\pi]\to\mathbb{R}$ be a continuous function. If $\int^{\pi}_{0}f (t) \sin(t)dt =\int^{\pi}_{0} f (t) \cos(t)dt = 0$, then $f(x)=0$ admits two solutions in $[0,\pi]$

I try to show if $f(x)>0$ and then get the contradiction but I failed to prove that, so maybe can someone help me with that? thanks in advance.

user10354138
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    I don't understand the last sentence. We can't have $f(x)>0$ on $[0,\pi]$ for then $f(x)\sin x>0$ on $(0,\pi)$ and $\int_0^\pi f(t)\sin t, \mathrm{d}t >0$ Similarly, we can't have $f(x)<0$ on $[0,\pi]$. Is this the contradiction you're speaking of? – saulspatz Feb 28 '21 at 16:06
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    Another one: https://math.stackexchange.com/q/247385/42969 – Martin R Feb 28 '21 at 16:09
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    @MartinR and another https://math.stackexchange.com/questions/3645891/given-int-0-pi-f-theta-cos-theta-d-theta-int-0-pi-f-theta-sin?noredirect=1&lq=1 although the answer was deleted – Albus Dumbledore Feb 28 '21 at 16:10
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    @AlbusDumbledore: The same answer was posted here: https://math.stackexchange.com/a/3646196/42969. Unfortunately, it is wrong. – Martin R Feb 28 '21 at 16:52

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