Finding roots: cosine function of x

[happyparticle]

Homework Statement

Finding roots cosine function of x

Relevant Equations

$\frac{-5}{4} +2cos(\frac{\pi d x}{Lv}) + cos(\frac{2\pi dx}{Lv}) = 0$

I need to find the zeros of this function where d,L,v are constants.
After several calculations I faced this equation.
I tried everything I know, but I can't solve this. Maybe I'm missing something or I must made a mistake earlier in the problem.
Thus, I would like to know if it is possible to find the roots analytically (wolfram gave me the roots).$\frac{-5}{4} +2cos(\frac{\pi d x}{Lv}) + cos(\frac{2\pi dx}{Lv}) = 0$

 

[anuttarasammyak]

By math rule of
cos 2kx= 2cos^2 kx-1
you can get quadratic equation of cos kx where $k=\pi d/Lv$

 

[happyparticle]

Brilliant!
However, since this is a cosine function I have the same answer for different x. I found that patern by guessing with the calculator. however, is there a way to find it mathematically.
I hope this is clear.

 

[anuttarasammyak]

However, since this is a cosine function I have the same answer for different x.

Please show us what you get actually for further investigation.

 

[FactChecker]

Brilliant!
However, since this is a cosine function I have the same answer for different x. I found that patern by guessing with the calculator. however, is there a way to find it mathematically.
I hope this is clear.

I think you missed his point. You can use @anuttarasammyak's equation to substitute and analytically solve a quadratic equation.

 

[happyparticle]

I good example is $cos(kx) + cos^2(kx) = -1/4$
$kx = 2.09$
However, $kx = 4.18, 8.36, 10.45, ...$ works aswell.

 

[anuttarasammyak]

Could you show "cos kx = .. " for your original OP and the new example in #6 by solving quadratic equations ?

 

[happyparticle]

For example in #6, I have $cos(kx) = \frac{-1 \pm \sqrt{1-1}}{2}$
$kx = arc cos( \frac{-1 \pm \sqrt{1-1}}{2})$

 

[anuttarasammyak]

1-1=0 so you can delete root. Arccos(-1/2) is easy to get.

 

[pasmith]

Brilliant!
However, since this is a cosine function I have the same answer for different x. I found that patern by guessing with the calculator. however, is there a way to find it mathematically.
I hope this is clear.

Remember that |\cos kx | \leq 1, so any solution of the quadratic equation which is outside this range can be discarded.

 

[happyparticle]

1-1=0 so you can delete root. Arccos(-1/2) is easy to get.

I mean, I still have the same question since arc cos(-1/2) $\approx$ 2.09.
However If I plug kx = 4.18 in #6, it works and |cos 4.18| < 1.

Is my question clear or I just don't see your point?

 

[anuttarasammyak]

I mean, I still have the same question since arc cos(-1/2) ≈ 2.09.

Draw a unit circle and the angle in it. You get precise radian or degree. That's what your teacher expects you to do, I suppose.

 

[happyparticle]

I understand this point. Where $x = {2\pi /3, 4\pi/3, 2\pi/3 + 2\pi + ...}$
however, I'm looking to have an expression for all the possible values of x.

 

[DrClaude]

$$\cos(\theta) = \cos(\theta + 2 n \pi), \quad \quad n \in \mathbb{Z}$$