Let f(x), x in [0, infty) be a non-negative c | vedclass

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(A) Given the inequality $f'(x) \cos x \le f(x) \sin x$ for $x \ge 0$.Rearranging the terms,we get $f'(x) \cos x - f(x) \sin x \le 0$.This expression

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(A) Given the inequality $f'(x) \cos x \le f(x) \sin x$ for $x \ge 0$.Rearranging the terms,we get $f'(x) \cos x - f(x) \sin x \le 0$.This expression is the derivative of the product $f(x) \cos x$ with respect to $x$,i.e.,$\frac{d}{dx}(f(x) \cos x) \le 0$.Let $g(x) = f(x) \cos x$. Since $g'(x) \le 0$,the function $g(x)$ is monotonically non-increasing for $x \ge 0$.For $x \in [0, \pi/2]$,$\cos x \ge 0$. As $x$ approaches $\pi/2$ from the left,$g(x) = f(x) \cos x$ must remain non-negative because $f(x) \ge 0$ and $\cos x \ge 0$.However,at $x = \pi/2$,$g(\pi/2) = f(\pi/2) \cdot 0 = 0$.Since $g(x)$ is non-increasing and $g(x) \ge 0$ for $x \in [0, \pi/2]$,and $g(\pi/2) = 0$,it implies $g(x) = 0$ for all $x \ge \pi/2$.Therefore,$f(2\pi) \cos(2\pi) = 0$,which means $f(2\pi) \cdot 1 = 0$,so $f(2\pi) = 0$.

Let $f(x), x \in [0, \infty)$ be a non-negative continuous function. If $f'(x) \cos x \le f(x) \sin x$ for all $x \ge 0$,then the value of $f(2\pi)$ is equal to

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