6. For any closed interval $I$, let $M_{1}$ denote the maximum value of the function $y=\sin x$ on $I$. If the positive number $a$ satisfies $M_{[0, a]}=2 M_{[a, 2 a]}$, then the value of $a$ is $\qquad$ .
Answer: $\frac{5}{6} \pi$ or $\frac{13}{12} \pi$.
Solution: Suppose $0 < a < \pi$, at this time $M_{[0, a]}=1$, so $M_{[a, 2 a]}=\frac{1}{2}$. Therefore, there exists a non-negative integer $k$, such that
$$
2 k \pi+\frac{5}{6} \pi \leq a<2 a \leq 2 k \pi+\frac{13}{6} \pi,
$$

and at least one of the two " $\leq$ " in (1) must be an equality.
When $k=0$, we get $a=\frac{5}{6} \pi$ or $2 a=\frac{13}{6} \pi$. Upon verification, $a=\frac{5}{6} \pi, \frac{13}{12} \pi$ both satisfy the conditions. When $k \geq 1$, since $2 k \pi+\frac{13}{6} \pi<2\left(2 k \pi+\frac{5}{6} \pi\right)$, there does not exist an $a$ that satisfies (1). In summary, the value of $a$ is $\frac{5}{6} \pi$ or $\frac{13}{12} \pi$.