Inspired by 3 SHS Bandung

$\frac { 36\sin ^{ 2 }{ x } +25 }{ \sin { x } } =\quad 36\sin x+\frac { 25 }{ \sin { x } }$

with $AM-GM$

$36\sin x+\frac { 25 }{ \sin { x } } \quad \ge \quad 2\sqrt { (36\sin { x } )(\frac { 25 }{ \sin { x } } ) } =2\sqrt {(36)(25)}=\boxed{60}$

and it could be happen when $\sin { x } =\frac { 5 }{ 6 }$

Let,

$f(x) = \dfrac{36 \sin^2 x + 25}{\sin x} = 36 \sin x + 25 \csc x$

Differentiating $f(x)$ w.r.t. $x$ we get,

$\begin{aligned} & 36 \cos x - 25 \csc x \cot x = 0 \\ \\ \implies & 36 \cos x - 25 \dfrac{\cos x}{\sin^2 x} = 0 \\ \\ \implies & \cos x \left( 36 - \dfrac{25}{\sin^2 x} \right) = 0 \\ \implies & \cos x = 0 ~~ \text{or} ~~ \sin x = \pm \dfrac{5}{6} \\ \implies & \sin x = \pm 1 ~~ \text{or} ~~ \sin x = \pm \dfrac{5}{6} \end{aligned}$

Since, $0 \le x \le \pi$ , therefore negative values of $\sin x$ are unacceptable.

Thus, we have $\sin x = 1$ or $\sin x = \dfrac{5}{6}$ . One of these values will give us the maximum value of the function and the other one the minimum value.

Checking, we get,

$\begin{aligned} \sin x = 1 & \implies & f(x) = 61 \\ \sin x = \dfrac 56 & \implies & f(x) = 60 \end{aligned}$

Thus, the minimum value of $f(x)$ is $\boxed{60}$ .