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.

By AM-GM, $4x + \dfrac{9\pi^2}{x} \ge 2\sqrt{4x \cdot \dfrac{9\pi^2}{x}} = 12\pi$, with equality iff $4x = \dfrac{9\pi^2}{x}$, i.e. $x = \dfrac{3\pi}{2}$. 
Also, $\sin x \ge -1$, with equality iff $x = \dfrac{3\pi}{2} + 2\pi k$ for some integer $k$. 
Therefore, $f(x) = 4x + \dfrac{9\pi^2}{x} + \sin x \ge 12\pi - 1$ with equality iff $x = \dfrac{3\pi}{2}$. 
Note: This solution does use the fact that the domain is $x > 0$, as stated in the question. 
