The domain of sin⁻¹⁡(3x) is equal to

The correct option is [−1/3, 1/3].

To find the domain of the function ${\sin }^{-1}\left(3x\right)$, we need to understand the domain of the standard inverse sine function, ${\sin }^{-1}\left(\theta \right)$.

The function ${\sin }^{-1}\left(\theta \right)$ is defined only when its input argument $\theta$ lies in the closed interval $[-1,1]$.
Therefore, we write the inequality:
$$-1\le \theta \le 1$$

For the given function ${\sin }^{-1}\left(3x\right)$, the argument is $3x$.
Applying the domain restriction to $3x$, we get:
$$-1\le 3x\le 1$$

To solve for $x$, we divide all parts of the inequality by $3$:
$$\frac{-1}{3}\le x\le \frac{1}{3}$$

This inequality represents all real numbers between $-1/3$ and $1/3$ inclusive.
Expressing this in interval notation, we get:
$$x\in [-1/3,1/3]$$

Thus, the domain of ${\sin }^{-1}\left(3x\right)$ is [−1/3, 1/3].