0
$\begingroup$

I have this equation

$$ 36 \cos \frac{3 x}{4} \cos \frac{27 x}{20} \left(\cos \frac{3 x}{5} +2 \cos \frac{21 x}{10} \right)=0 $$

Is it possible to ask Mathematica to give all the roots of the function on the domain $0<x<4\pi$? Preferably as a rational multiple of $\pi$? (the plot of the function is attached)

36 Cos[(3 x)/4] Cos[(27 x)/20] (Cos[(3 x)/5] + 2 Cos[(21 x)/10]) == 0

enter image description here

$\endgroup$

1 Answer 1

2
$\begingroup$
sol = x /. Solve[36 Cos[(3 x)/4] Cos[(27 x)/20] (Cos[(3 x)/5] + 2 Cos[(21 x)/10]) == 0&& 0 < x < 4 Pi]

Plot[36 Cos[(3 x)/4] Cos[(27 x)/20] (Cos[(3 x)/5] + 
    2 Cos[(21 x)/10]), {x, 0, 4 Pi}, 
 Epilog -> {Red, 
   Point[Transpose[{sol, ConstantArray[0, Length@sol]}]]}]

enter image description here

enter image description here

If you desire only roots that are rational multiples of Pi, you can use (among many other ways):

Select[sol, #/Pi \[Element] Rationals &]
$\endgroup$
3
  • $\begingroup$ Thanks, Then, does this mean that those $ArcTan[...]$ are definitely not multiple of $\pi$? Is it correct to use $N[ArcTan ...]$ to know their numerical value? $\endgroup$
    – user67849
    Apr 15, 2021 at 0:47
  • 1
    $\begingroup$ @quark - it means they're not rational multiples of Pi, and yes, N will get you the numeric result. $\endgroup$
    – ciao
    Apr 15, 2021 at 0:48
  • $\begingroup$ Strictly speaking, this doesn't prove that the ArcTan[..] terms are not rational multiples of Pi. They may be. $\endgroup$ Apr 15, 2021 at 2:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.