Simplify the absolute value 3|\cos ?| if ? = \sin^{−1} \frac{x}{3}

Step 1
Given angle is,
${\displaystyle 0={\sin }^{-1}\frac{x}{3}}$
Here, the objective is to find the simplify for absolute ${\displaystyle {\sin }^{-1}\frac{x}{3}}$.
${\displaystyle \sin 0=\frac{x}{3}}$
Then, by the trigonometric rule.
${\displaystyle \cos 0=\sqrt{1-{\left(\frac{x}{3}\right)}^2}}$
${\displaystyle =\sqrt{1-\frac{x^2}{9}}}$
${\displaystyle =\sqrt{\frac{9-x^2}{9}}}$
${\displaystyle =\frac{1}{3}\sqrt{9-x^2}}$
Step 2
Then, the absolute value of ${\displaystyle 3\left|\cos 0\right|}$.
${\displaystyle 3\left|\cos 0\right|=3\left|\frac{1}{3}\sqrt{9-x^2}\right|}$
${\displaystyle =\sqrt{9-x^2}}$
Thus, this the absolute value.