We need to solve the inequality $\sqrt{(a-3)^2} \sin a - 3 < 0$.

1. Problem Description

We need to solve the inequality $\sqrt{(a-3)^2} \sin a - 3 < 0$.

2. Solution Steps

First, we simplify the square root:

$\sqrt{(a-3)^2} = |a-3|$.

So the inequality becomes $|a-3| \sin a - 3 < 0$, which can be rewritten as $|a-3| \sin a < 3$.

Now, we consider two cases:

Case 1: $a \ge 3$. In this case, $|a-3| = a-3$, so the inequality becomes $(a-3) \sin a < 3$.

Case 2: $a < 3$. In this case, $|a-3| = 3-a$, so the inequality becomes $(3-a) \sin a < 3$.

Now, let us consider the expression $|a-3|\sin a$. Since $-1 \le \sin a \le 1$, then

$-|a-3| \le |a-3|\sin a \le |a-3|$.

So, if $|a-3| \le 3$, then $|a-3| \sin a \le 3$.

We have $|a-3| \le 3$, which implies $-3 \le a-3 \le 3$.

Adding 3 to all parts of the inequality, we have $0 \le a \le 6$.

If $a=3$, then $|a-3| = 0$, so $|a-3|\sin a = 0 < 3$. Thus $a=3$ is a solution.

If $a=0$, then $|a-3| = 3$, and $\sin a = \sin 0 = 0$, so $|a-3|\sin a = 0 < 3$. Thus $a=0$ is a solution.

If $a=6$, then $|a-3| = 3$. $\sin a = \sin 6 \approx -0.2794$, so $|a-3|\sin a = 3 \sin 6 \approx -0.838 < 3$. Thus $a=6$ is a solution.

If $a > 6$, then $|a-3| = a-3 > 3$. Also, we know that $\sin a \le 1$, so $|a-3| \sin a$ could be less than

3. If $a$ is close to $2\pi$, say $a = 2\pi \approx 6.28$, then $|a-3| = |6.28-3| = 3.28$. $\sin a = \sin 2\pi = 0$, so $|a-3|\sin a = 0 < 3$. However, as $a$ gets larger, it is difficult to determine whether the inequality holds in general.

Let's analyze when $(a-3) \sin a < 3$ and $(3-a) \sin a < 3$ are not satisfied. This occurs when $a>3$ and $\sin a > \frac{3}{a-3}$ or when $a<3$ and $\sin a > \frac{3}{3-a}$.

When $\sin a < 0$, we are guaranteed that the inequalities hold.

When $\sin a = 0$, the inequalities are satisfied for any $a$.

Consider the original inequality $|a-3| \sin a < 3$.

This inequality is satisfied when $\sin a < 0$ or when $|a-3| < \frac{3}{\sin a}$ when $\sin a > 0$.

3. Final Answer

$0 \le a \le 6$