TITLE: How to find if this function is always zero?
QUESTION [12 upvotes]: The question I faced was:

Let $f(x)$ be a non-negative continuous and bounded function for all $x \ge 0$. If
  $$(\cos x)f'(x) \le (\sin x - \cos x)f(x), \; \forall \; x \ge 0$$
  then which of the following is/are correct?
(A) $f(6) + f(5) > 0$
(B) $x^2 - 3x + 2 + f(9) = 0$ has two distinct solutions
(C) $f(5)f(7) - f(6)f(5) = 0$
(D) $\lim\limits_{x \to 4} \dfrac{f(x) - \sin(\pi x)}{x-4} = 1$

The answer is:

 (B), (C)

By observation, $f(x)=0$ satisfies the given conditions. But is it the only solution? If so, how to prove it is?
After rearranging the terms and combining them, I converted the inequality to this form:
$$ \left( f(x)\,\cos x \right)' + f(x)\,\cos x \le 0 $$
Despite its allure, this inequality isn't getting me anywhere! It doesn't seem to have any information about $f(x)$, since it is stuck with a "$\cos x$". Even then, I don't see where I can go with it.
So how to solve this problem? Thank you.

REPLY [3 votes]: Notice that whenever some $f$ satisfies the hypothesis, then so does any positive multiple of $f$. In particular:
Note that (B) is equivalent to
$$0<(-3)^2-4(2+f(9)) = 1-4f(9)$$
i.e. $f(9)<1/4$. Hence (B) can hold if and only if $f(9) =0$ for any $f$ satisfying the hypothesis.