Find the minimum value of $\frac{9x^2\sin^2 x + 4}{x\sin x}$ for $0 < x < \pi.$
I notice that the expression $\frac{9x^2\sin^2 x + 4}{x\sin x}$ can be rewritten as $9x\sin x + \frac{4}{x\sin x}.$ This suggests that I might want to use the AM-GM inequality, which says that for any nonnegative numbers $a$ and $b,$ the arithmetic mean $\frac{a+b}{2}$ is greater than or equal to the geometric mean $\sqrt{ab}.$ Applying this inequality to $9x\sin x$ and $\frac{4}{x\sin x},$ I get that $\frac{9x\sin x + \frac{4}{x\sin x}}{2} \geq \sqrt{9x\sin x \cdot \frac{4}{x\sin x}} = \sqrt{36} = 6.$ Multiplying both sides by 2, I get that $9x\sin x + \frac{4}{x\sin x} \geq 12.$ This means that the minimum value of the original expression is 12, and it occurs when the AM-GM inequality becomes an equality, which happens when $9x\sin x = \frac{4}{x\sin x}.$ Solving for $x,$ I get that $x\sin x = \frac{2}{3}.$ I use a graphing calculator and find that the solution is approximately $x \approx 0.878.$ 