Local minimum of $f(x) = 4x + \frac{9\pi^2}{x} + \sin x$ What's the minimum value of the function
$$f(x) = 4x + \frac{9\pi^2}{x} + \sin x$$
for $0 < x < +\infty$? The answer should be $12\pi - 1$, but I get stuck with the expression involving both $\cos x$ and $x^2$ in the derivative. Taking the derivative, we have:
$$f'(x) = 4 - \frac{9\pi^2}{x^2} + \cos x.$$
In order to find the local extrema of the function, we set $f'(x) = 0$. Therefore,
\begin{align}
4 - \frac{9\pi^2}{x^2} + \cos x &= 0 \\
4x^2 - 9\pi^2 + x^2 \cos x &= 0 \\
x^2 (4 + \cos x) &= 9\pi^2.
\end{align}
However, I'm not sure what to do from here or if, indeed, I'm doing it right at all. Any help would be appreciated.

Let's try breaking the equation into two parts; I'll explain below.
So we have $\displaystyle f'(x)= \underbrace{4-\frac{9\pi^2}{x^2}}_{0}+\underbrace{\cos x}_{0}=0$. Note the underbraces; I set the terms equal to $0$ (so we can have $0+0=0$):
$$4-\frac{9\pi^2}{x^2}=0 \text{ and } \cos x =0$$
From $\displaystyle4-\frac{9\pi^2}{x^2}=0$, we can algebraically solve it and get $\displaystyle x=\pm\frac{3\pi}{2}$. Also, $\displaystyle \cos \left( \pm\frac{3\pi}{2} \right)=0$.
Putting it altogether, at $\displaystyle x=\pm\frac{3\pi}{2}$,
$$\left( 4-\frac{9\pi^2}{x^2} \right)+(\cos x)=0+0=0$$
which verifies. 
(This isn't the best way to solve it, but it's a lot like trying to solve $x+\sin x = 0$. You can't solve it analytically, but we can easily guess $x=0$ is the solution because $0+\sin 0 = 0+0=0$.)
Back to your problem. If we have $\displaystyle x=\pm\frac{3\pi}{2}$, well we can drop the negative value and just have $\displaystyle x=\frac{3\pi}{2}$. Why? Well the domain stated in your problem, $0 < x < \infty$, allows only positive real numbers.
Now you can plug in $\displaystyle \frac{3\pi}{2}$ into your $f(x)$. You will get $f\left( \frac{3\pi}{2} \right)=12\pi - 1$ as your answer.
