5. Given the equation about $x$: $(2 a-1) \sin x+(2-a) \sin 2 x=\sin 3 x$, if all solutions of the equation in the range $x \in[0,2 \pi]$ form an arithmetic sequence in ascending order, then the range of the real number $a$ is $\qquad$ .
$$
\begin{array}{l}
(2 a-1) \sin x+2(2-a) \sin x \cos x=3 \sin x-4 \sin ^{3} x \\
\Rightarrow \sin x\left[2 \cos ^{2} x+(a-2) \cos x-a\right]=\sin x(\cos x-1)(2 \cos x+a)=0 .
\end{array}
$$

Notice that $\sin x(\cos x-1)=0$ has all roots $0, \pi, 2 \pi$ in $[0,2 \pi]$, forming an arithmetic sequence; when $a=0$, $\cos x=0$ has all roots $\frac{\pi}{2}, \frac{3 \pi}{2}$ in $[0,2 \pi]$, and at this time, these 5 numbers can form an arithmetic sequence; when $a \in(-2,0) \cup(0,2)$, the two roots of $\cos x=-\frac{a}{2}$ in $[0,2 \pi]$ cannot form an arithmetic sequence with $0, \pi, 2 \pi$. In summary, the range of real number $a$ is $(-\infty,-2] \bigcup\{0\} \bigcup[2,+\infty)$.