$ \int_0^\pi f (x) \sin x=\int_0^\pi f (x) \cos x=0$.Prove that $f (x) $ has two roots in the interval $(0, \pi).$

Asked Nov 24 '16 at 02:37

Active Nov 24 '16 at 02:37

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f is a continuous function. $ \int_0^\pi f (x) \sin x=\int_0^\pi f (x) \cos x=0 $.Prove that $f (x)$ has two roots in the interval $(0 ,\pi)$

Look that $\sin x>0$ over the whole interval and thus to make the integral $0$, we must have $f(x)<0$ at some point.But if we get a root, we may transform the origin to the root and then use the sign changing property of $\cos x$

asked Nov 24 '16 at 02:37

What is your question? – John Hughes Nov 24 '16 at 02:44
Prove that $f(x)$ has 2 roots in $(0,\pi)$. – Nov 24 '16 at 02:46
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Duplicate of Prove that $ f(x) $ has at least two real roots in $ (0,\pi) $ – Momo Nov 24 '16 at 02:59
Note: that's not a question, it's a demand. Less likely to get results in general. :) – John Hughes Nov 24 '16 at 03:06

$ \int_0^\pi f (x) \sin x=\int_0^\pi f (x) \cos x=0$.Prove that $f (x) $ has two roots in the interval $(0, \pi).$

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