Extreme points of $f(t)=h(t)\cos(t)+g(t)\sin(t)$ Let $f:[0,12]\to R$ , $f(x)=3x\cos(x)+2x^3\sin(x)$ and I have to find $\min(f(x))$ and $\max(f(x))$. I have tried the derivative  and some other things like observing the minimum is between $3\pi$ and $\left(3+\frac12\right)\pi$ but cant find the exact value. I have tried to see if any formula for $a\cos(t)+b\sin(t)$ where $a$, $b$ are fixed constants may work here but it's not the case (I tried Cauchy–Bunyakovsky–Schwarz inequality and the so called R-method) 
PS: first post here so excuse me if i break the rules in any way.
 A: Using what you observed cocerning the approximate location, we can approximate the solution expanding the derivative
$$f'(x)=\left(2 x^3+3\right) \cos (x)+3 x (2 x-1) \sin (x)$$ as Taylor series built around $x=\frac 72\pi$. Making $x=t+\frac 72\pi$, this would give
$$\left(\frac{21 \pi }{2}-\frac{147 \pi ^2}{2}\right)+\left(6-42 \pi +\frac{343 \pi
   ^3}{4}\right) t+\left(-6-\frac{21 \pi }{4}+\frac{441 \pi ^2}{4}\right)
   t^2+O\left(t^3\right)$$
Using the expansion to $O\left(t^2\right)$ would give
$$t=\frac{42 \pi  (7 \pi -1)}{24-168 \pi +343 \pi ^3}\implies x\approx 11.2690$$
Using the expansion to $O\left(t^3\right)$  would give $x\approx  11.2432$.
What we could also it to use the simplest $[1,1]$ Padé approximant which would lead to
$$t=\frac{42 \pi  (7 \pi -1) \left(24-168 \pi +343 \pi ^3\right)}{576-7056 \pi +22050   \pi ^2-8232 \pi ^3+14406 \pi ^4+117649 \pi ^6}$$ which implies $x\approx 11.2408$.
Starting with any of these estimates, Newton method will converge in very few iterations. 
$$\left(
\begin{array}{cc}
 n & x_n \\
 0 & 11.26895470 \\
 1 & 11.24507394 \\
 2 & 11.24492998
\end{array}
\right)$$
A: By using the so called r-formula, to which you already refer, you will see, that there is no global minimum or maximum (except the obvious $\pm\infty$). The extrema of the function you gave cannot be given in a closed form, you will always end up with a transcendental equation of the form $\sin(x\pm \arctan(\frac{3}{2x^2}))=\pm 1$ that you can only solve by numerical methods. The first (local) minimum is indeed located at $x=10.9831407366212\dots$.
Hope this is of any use to you. If you'd like more details, I can elaborate on that later, let me know.
Apology: I am terribly sorry, but Claude Leibovici is absolutely correct! While scribbling on a small piece of paper, I lost a term which should not have been lost. Sorry again!
