Does $\int_R \cos(rx) f(x) = 0$ with $f(x)=f(-x)$ imply $f(x) = 0$ almost everywhere?

Question

Assume $f(x)\in L_1$, that is $\int_R |f(x)|dx <\infty$. Easy to know $\int_R \cos(rx)f(x)dx$ is well-defined. My question is if $\int_R \cos(rx) f(x) = 0$ for any $r\geq 0$, can we come to be conclusion that $f(x)=0$ almost everywhere? (Some people say $f(x)$ being an odd function will be a counterexample. Let's further assume $f(x)$ to be an even function.)

Answer 1

If $f$ is even, then $$ \hat f(r)=\int_{\Bbb R}f(x)\,e^{irx}\,dx=\int_{\Bbb R}f(x)\,\cos(r\,x)\,dx=0\quad\forall r\in\Bbb R. $$ By the uniqueness theorem for Fourier transforms, $f$ is equal to zero almost everywhere.

Answer 2

No. Let $f$ be any integrable odd function, such as $f(x)=x\exp(-x^2)$.