Question
Let $f\left( x \right), x \in \left[ {0,\infty } \right)$ be a non-negative continuous function. If $f'\left( x \right)\cos x \le f\left( x \right)\sin x\ \forall\, x \ge 0$, then the value of $f(2\pi)$ is equal to
$f^{\prime}(x) \cos x-f(x) \sin x \leq 0$
$(f(x) \cos x)^{\prime} \leq 0$
$\therefore $ $g(x)=f(x) \cos x$ is a monotonically nonincreasing function.
Since $\mathrm{f}(\mathrm{x})$ is non-negative, $\mathrm{f}(2 \pi)=0$
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