How to find if this function is always zero?

If you set $g(x)=f(x)\cos x$, your rearrangement tells you that $$ \tag{1} g'(x) \le -g(x) $$

Consider an interval $[2\pi k-\frac12\pi , 2\pi k +\frac12\pi]$ where $\cos x$ is $\ge 0$. Then you know that $g(x)\ge 0$ on this interval, and $g(x)=0$ at the start of the interval. Then, because of (1), $g(x)$ must be identically zero on that interval, and so must $f(x)$.

This settles at least (A) and (C), because $5$, $6$ and $7$ are all within $\pi/2$ of $2\pi$.

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For (B) and (D) this argument doesn't tell you enough; I recommend Peter's slicker answer instead.

We assume that $f(x)$ is bounded and non-negative, for $x \geqslant 0$. This means that there is a non-negative function $g(x)$ with the property that $f(x) = g(x)e^{-x}$ for $x \geqslant 0$.

Plugging this into the inequality you found gives, \begin{equation} 0 \geqslant (f(x)\cos (x))' + f(x) \cos (x) = e^{-x} (g(x) \cos(x))'. \end{equation} And since $e^{-x} > 0$, we have \begin{equation} 0 \geqslant (g(x) \cos(x))'. \end{equation} This means that the function $g(x) \cos(x)$ is weakly decreasing. Because any point $x \in \mathbb{R}$ is between two zeroes of $\cos(x)$, we have that $g(x) \cos(x) = 0$ for all $x$.

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.