How to solve $ \tan (m \theta) + \cos(n \theta) =0$ $$ \tan (2m \theta) + \cos(2n \theta) =0 $$
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 ?
 A: 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 $$t_{n+1}=t_n+\frac{\cos (t_n) \cos (r t_n)+\sin (t_n)}{\sin (t_n) \cos (r t_n)+\cos (t_n) (r \sin (r t_n)-1)}$$
A: 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.
