Find the range of the function $f(x)=3\sin\left(\sqrt{\frac{π^2}{16}-x^2}\right)$.

$\left[0,\frac{3}{\sqrt{2}}\right]$

We must have $\frac{π^2}{16}-x^2≥0$

Also $\frac{π^2}{16}-x^2≤\frac{π^2}{16}$

$⇒0≤\frac{π^2}{16}-x^2≤\frac{π^2}{16}$

$⇒0≤\sqrt{\frac{π^2}{16}-x^2}≤\frac{π}{4}$

$⇒0≤\sin\sqrt{\frac{π^2}{16}-x^2}≤\frac{1}{\sqrt{2}}$

$⇒0≤3\sin\sqrt{\frac{π^2}{16}-x^2}≤\frac{3}{\sqrt{2}}$

Hence, range is $\left[0,\frac{3}{\sqrt{2}}\right]$