$$ \tan (2m \theta) + \cos(2n \theta) =0 $$

I am trying to solve this trigonometric equation of theta to get a general solution.

I tried this question by using the substitutions $t_1= \tan(n\theta)$ and $t_2= \tan(m \theta)$.

But i was not able to get a good answer.

Any ideas ?

  • $\begingroup$ If $\frac nm$ is a whole number, we may employ chebyshev polynomials to see this reduces down to factoring a polynomial. $\endgroup$ – Simply Beautiful Art Jan 8 '17 at 1:29
  • $\begingroup$ And clearly if we get polynomials degree $5$ or higher, I cannot expect easily accessible solutions. $\endgroup$ – Simply Beautiful Art Jan 8 '17 at 1:30

As marty cohen answered, for the most general case, only numerical methods will be able to solve the equation $$\tan (2m \theta) + \cos(2n \theta) =0$$ which, as he wrote, would be better written as $$\tan(t)+\cos(rt)=0$$ The problem is that the function $$f(t)=\tan(t)+\cos(rt)$$ presents an infinite number of discontinuities at $t=(2k+1)\frac \pi 2$ and thius is never very good.

Assuming $\cos(t)\neq 0$ as a possible solution, I suggest that you look instead for the zero's of $$g(t)=\sin(t)+\cos(t)\cos(rt)$$ which is continuous everywhere.

Starting froma "reasonable" guess $t_0$, Newton method will updtate it according to $$t_{n+1}=t_n+\frac{\cos (t_n) \cos (r t_n)+\sin (t_n)}{\sin (t_n) \cos (r t_n)+\cos (t_n) (r \sin (r t_n)-1)}$$

  • $\begingroup$ I have a feeling that the same person downvoted both of us. $\endgroup$ – marty cohen Jan 8 '17 at 6:11
  • $\begingroup$ @martycohen. I do think so ! Cheers. $\endgroup$ – Claude Leibovici Jan 8 '17 at 6:12

I think that it is unlikely that there is a general solution.

By letting $t = 2m\theta$ and $r = n/m$, $\tan (2m \theta) + \cos(2n \theta) =0 $ becomes $\tan (t) = \cos(rt) $.

Looking at the graphs of $\tan(t)$ and $\cos(rt)$, we see that there is at least one solution in each range $k\pi \le t < (k+1)\pi$. Therefore there are a countable infinity of solutions independent of the values of $n$ and $m$.

Also, if $r < 1$ (or $n < m$), there is exactly one root in each interval.

So it seems to me that only numerical calculations can find the roots in general.

However, for small values of $n$ or $m$, there may be explicit solutions.


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