How do the coefficients in the linear combination of cosines impact the number of local minima of the sum? Consider the following function: $$f(\theta) =  r_0 + r_1 \cos(\theta + \phi_1) + r_2 \cos(2\theta + \phi_2)$$ where $\theta$ is an angle between 0 and $2\pi$. For all $0\leq k\leq 2$ we have $r_k\geq 0$ and $0\leq \phi_k \leq 2\pi$, and that the tuple $\{r_k,k\theta + \phi_k\}$ gives the amplitude and phase for the $k$th cosine in the linear combination.
Observations.
I have observed that when $r_2 \geq r_1/3$ (a crude, empirical observation) the shape of $f(\theta)$ always has two local minima regardless of the phase shifts $\phi_1, \phi_2$. 
On the other hand, when $r2 \leq r_1/5$ (again, a crude observation) $f(\theta)$ has only one minimum. The variation in phase shifts does not change the number of minima of the sum although it changes the shape of the function.
I appears that if one of the $r_k$s is "much" large compared to the other $r_k$s then the cosine corresponding to the only large $r_k$ will dictate the shape of the sum. 
It also appears that when the sum has two minima, they always appear "close" to the two minima of $\cos(2\theta)$. 
Question.
Is it possible to tell, by looking at the values of $r_k$, the number of local minima that $f(\theta)$ will have? In particular, can we determine a threshold condition when the sum goes from having only one minimum to two minima (and vice versa)?
I think the answer should already be known. What are some known results / resources related to this observation?
Mathematica script.
Here is a simple mathematica script for visualizing the linear combination of three cosines. The blue curve corresponds to the sum. Setting $r_3=0$ in the interactive plot would correspond to the case discussed above.
f[x_, r0_, r1_, phi1_, r2_, phi2_, r3_, phi3_] :=  
 Re[r0 + r1  Exp[I (x + phi1) Pi] + r2  Exp[I (2 x + phi2) Pi] + 
   r3 Exp[I (3 x + phi3) Pi]]


Manipulate[
 Plot[{f[x, r0, r1, phi1, r2, phi2, r3, phi3], r0, 
   Re[r1 Exp[I (x + phi1) Pi]], Re[ r2 Exp[I (2 x + phi2) Pi]], 
   Re[r3 Exp[I (3 x + phi3) Pi]]}, {x, 0, 
   2}, {PlotStyle -> blue}], {r0, 0, 100}, {r1, 0, 100}, {phi1, 0, 
  2}, {r2, 0, 100}, {phi2, 0, 2}, {r3, 0, 100}, {phi3, 0, 2}]

Update. According to the Wikipedia entry about "Trigonometric polynomials",  our function $f(\theta)$ is a trigonometric polynomial of degree 2 and hence can have at most 4 roots. Because $f(\theta)$ is also periodic, we will have at most 2 (local) maxima and minima. 
How can we characterize the locations of these critical points in terms of the coefficients $\{r_k, \phi_k\}$? Or possibly only $\{r_k\}$ since apparently the phase shifts do not to move the critical points beyond a certain limit. What is known so far?
 A: If you plot in polar coordinates $$
\left\{ \matrix{
  x(\theta ) = r_{\,1} \cos \left( {\theta  + \phi _{\,1} } \right) + r_{\,2} \cos \left( {2\theta  + \phi _{\,2} } \right) \hfill \cr 
  y(\theta ) = r_{\,1} \sin \left( {\theta  + \phi _{\,1} } \right) + r_{\,2} \sin \left( {2\theta  + \phi _{\,2} } \right) \hfill \cr}  \right.
$$
then it is clear that only the ratio of the $r$'s matters and that the difference of the phases rotates the whole diagram. 
Let then consider
$$
\eqalign{
  & \left( {f(\theta ) - r_{\,0} } \right)/r_{\,1}  = \cos \left( {\theta  + \phi _{\,1} } \right) + r_{\,2} /r_{\,1} \cos \left( {2\theta  + \phi _{\,2} } \right) =   \cr 
  &  = \left( {\cos \left( {\theta  + \phi _{\,2} /2 + \phi _{\,1}  - \phi _{\,2} /2} \right) + r_{\,2} /r_{\,1} \cos \left( {2\left( {\theta  + \phi _{\,2} /2} \right)} \right)} \right) \cr} 
$$
and so let's transform that for simplicity into:
$$
g(\alpha ) = \cos \left( {\alpha  + \delta } \right) + \rho \cos \left( {2\alpha } \right)
$$
from which:
$$
\eqalign{
  & 0 = {d \over {d\alpha }}g(\alpha ) =  - \sin \left( {\alpha  + \delta } \right) - 2\,\rho \sin \left( {2\alpha } \right)  \cr 
  &  =  - \cos \delta \sin \alpha  - \sin \delta \cos \alpha  - 4\,\rho \sin \alpha \cos \alpha  =   \cr 
  &  =  \mp \cos \delta {{\tan \alpha } \over {\sqrt {1 + \tan ^{\,2} \alpha } }} \mp \sin \delta {1 \over {\sqrt {1 + \tan ^{\,2} \alpha } }} - 4\,\rho {{\tan \alpha } \over {1 + \tan ^{\,2} \alpha }} \cr} 
$$
which, leaving $\delta$ generic, leads to a 4th degree equation.
Taking instead $\delta=0$, the polar plot is symmetric around the $x$ axis, and we can easily solve into
$$
{d \over {d\alpha }}g(\alpha ) = 0\quad \left| {\;\delta  = 0} \right.\quad  \Rightarrow \quad \sin \alpha \left( {1 + 4\,\rho \cos \alpha } \right) = 0\quad  \Rightarrow \quad \left\{ \matrix{
  \sin \alpha  = 0\;\;\left( {{\rm maxima}} \right) \hfill \cr 
  \cos \alpha  =  - 1/\left( {4\,\rho } \right)\;\;\left( {{\rm minima}} \right) \hfill \cr}  \right.
$$
which means that for $\rho  < 1/4\,$ the minima disappears, while for $\rho  >  > 1\,$ they get close to $\alpha  =  \pm \;\pi /2$
