If $(\cos x)f'(x)\leq (\sin x-\cos x)f(x)$ for all $x\geq 0$. Can it be said that $f(x)$ is a constant. Let $f(x)$ be a non-negative continuous and bounded function for all $x\geq 0$.
If $$(\cos x)f'(x)\leq (\sin x-\cos x)f(x)$$ for all $x\geq 0$.
Can it be said that $f(x)$ is a constant.
My Attempt
$(\cos x)f'(x)\leq (\sin x-\cos x)f(x)$
$\cos x(f'(x)+f(x))+(-\sin x)f(x)\leq 0$
$\cos x e^x(f'(x)+f(x))+(-\sin x)e^xf(x)\leq 0$
Thus $\frac{d}{dx}(e^x\cos xf(x))\leq 0$
So, for $x\geq 0$ we have
$e^x\cos xf(x)\leq f(0)$
Here I am stuck as the value of $f(0)$ is not supplied. Can it be deduced that $f(0)=0$

As you correctly said, the condition is equivalent to $h(x) = e^x \cos(x) f(x)$ being (weakly) decreasing on $[0, \infty)$.
Since $h(\pi/2 + k\pi) = 0$ for all non-negative integers $k$, it follows that $h$ is zero for $x \ge \pi/2$, which in turn implies that $f$ is zero for $x \ge \pi/2$.
We can also conclude that if $f(x_0) = 0$ for some $x_0 \ge 0$ then $f(x) = 0$ for all $x \ge x_0$.
But $f$ can be non-zero on an initial interval: If we choose an arbitrary twice differentiable function $h$ which is strictly decreasing on $[0, \pi/2]$ with $h(\pi/2) = h'(\pi/2) = h''(\pi/2) = 0$  then
$$
 f(x) = \begin{cases}
\frac{h(x)}{e^x \cos(x)} & 0 \le x < \frac \pi 2 \\
 0 & x \ge \frac \pi 2
\end{cases}
$$
is differentiable, satisfies $\frac{d}{dx}(e^x\cos xf(x))\leq 0$ for all $x \ge 0$, but is non-zero on $[0, \pi/2)$.
