7. Given that the function $f(x)$ satisfies $f(x)+f(-x)=0$ on $R$, and when $x>0$, $f(x)=\frac{1}{2}(|x+\sin \alpha|+|x+2 \sin \alpha|)+\frac{3}{2} \sin \alpha$. If for any $x \in R, \quad f(x-3 \sqrt{3}) \leq f(x)$ always holds, then the range of $\alpha$ is $\qquad$
7. $2 k \pi-\frac{\pi}{3} \leq \alpha \leq 2 k \pi+\frac{4}{3} \pi, k \in Z$

Analysis: Let $t=\sin \alpha$, then $t \in[-1,1]$; Since $x>0$, $f(x)=\frac{1}{2}(|x+t|+|x+2 t|)+\frac{3}{2} t$; If $t \geq 0$, then when $x>0$, $f(x)=x+3 t$, and when $x<0$, $f(x)=-f(-x)=x-3 t$; From $f(x-3 \sqrt{3}) \leq f(x)$ always holding, it can be deduced that the graph of $y=f(x)$ is always above the graph of $y=f(x-3 \sqrt{3})$, thus $\sin \alpha \geq 0$; If $t<0$, when $x \geq 0$, $f(x)=\left\{\begin{array}{l}-x, x \leq-t \\ t,-t<x<-2 t \\ x+3 t, x \geq-2 t\end{array}\right.$, from $f(x)=x+3 t, x \geq-2 t$ we get $f(x) \geq t$; When
$-t<x<-2 t$, we get $f(x)=t$; From $f(x)=-x, 0 \leq x \leq-t$ we get $f(x) \geq t$. Therefore, when $x \geq 0$, $f(x)_{\min }=t$. Since $f(x)$ is an odd function, when $x<0$, $f(x)_{\max }=-t$, and for $x \in R$, $f(x-3 \sqrt{3}) \leq f(x)$ always holds, then $-3 t-3 t \leq 3 \sqrt{3}$, which gives $t \geq-\frac{\sqrt{3}}{2}$, i.e., $\sin \alpha \geq-\frac{\sqrt{3}}{2}$, solving for $\alpha$ gives the range $2 k \pi-\frac{\pi}{3} \leq \alpha \leq 2 k \pi+\frac{4}{3} \pi$.