11. (15 points) Let the function $f(x)=\frac{\sin x}{2+\cos x}$. If for any $x^{3} \geqslant 0$, we have $f(x) \leqslant a x$, find the range of the real number $a$.

保留源文本的换行和格式，直接输出翻译结果。
11. Solution: Let $g(x)=a x-f(x)$,

then $g^{\prime}(x)=a-\frac{2 \cos x+1}{(2+\cos x)^{2}}=a-\frac{2}{2+\cos x}+ \frac{3}{(\cos x+2)^{2}}=3\left(\frac{1}{2+\cos x}-\frac{1}{3}\right)^{2}+a-\frac{1}{3}$,

thus when $a \geqslant \frac{1}{3}$, $g^{\prime}(x) \geqslant 0$, so $g(x)$ is an increasing function on $[0, \infty)$, and since $g(0)=0$, when $x \geqslant 0$, $g(x) \geqslant g(0)=0$, i.e., $f(x) \leqslant a x$.

When $0 < a < \frac{1}{3}$, let $h(x)=\sin x - 3ax$,

then $h^{\prime}(x)=\cos x - 3a > 0$, so $h(x)$ is an increasing function on $(0, \arccos 3a)$, hence when $x \in (0, \arccos 3a)$, $h(x) > h(0) = 0$, i.e., $\sin x > 3ax$.

Thus, when $x \in (0, \arccos 3a)$, $f(x)=\frac{\sin x}{2+\cos x} > \frac{\sin x}{3} > ax$.

When $a < 0$, $f(x) > 0 \geqslant a \cdot \frac{\pi}{2}$, therefore, the range of $a$ is $\left[\frac{1}{3},+\infty\right)$.