# Solving a Trigonometric Equation

1. Apr 27, 2005

### Night Owl

How would you solve for $$\alpha$$ in the following equation?

$$4=\cos(\alpha)+\cos^2(\alpha)+\cos^4(\alpha)$$

2. Apr 28, 2005

### whozum

Maple says there are no roots.

3. Apr 28, 2005

### Night Owl

Really? :( That would make me a very sad owl. Why are there no roots? I haven't heard of this "Maple" character yet...! ;) I'm only in Math 126 Honors (current material includes things ranging from taylor polynomials, maclaurin series, vectors in three-space, parametric equations for all sorts of stuff). Why is it not possible to solve for alpha? You could solve the following for at least one case:

$$3 = \cos\alpha + \cos^2\alpha + \cos^4\alpha$$

$$\alpha = 0$$

That works, doesn't it? I mean, all you do is think to yourself and say:

$$\cos0 = 1$$

And then you're golden, because any power of cosine is then also 1, so if you add three cosines raised to different powers, you'll get 1 + 1 + 1 = 3. So it seems reasonable that there might be a way to solve for alpha in the previous case.

Last edited: Apr 28, 2005
4. Apr 28, 2005

### whozum

arccos(RootOf(_Z+_Z^2+_Z^4-4,index = 1)),arccos(RootOf(_Z+_Z^2+_Z^4-4,index = 2))

Unless you consider those roots.

5. Apr 28, 2005

### Night Owl

Err....what? I don't understand that expression.

6. Apr 28, 2005

### Moo Of Doom

There cannot be a solution to that equation in the reals because the max of cos(x) is 1, and thus the max of cos^2(x) and cos^4(x) are also 1. The highest value of cos(x)+cos^2(x)+cos^4(x) is 3. I also believe 0 is the lowest value for it.

7. Apr 28, 2005

### Night Owl

Huh. I see. Kind of like how you can't solve for -1 = x^2 using the reals alone, maybe... But I mean, suppose that you had, in general

$$a = \cos\theta + \cos^2\theta + \cos^4\theta$$

Supposing that "a" were a number for which you could find a theta which would satisfy the equation, how would you go about solving it?

Last edited: Apr 28, 2005
8. Apr 28, 2005

### James R

Using Maple?

9. Apr 28, 2005

### whozum

You could try factoring. Maple would be your best bet, but graphing the function wouldnt hurt.

10. Apr 28, 2005

### dextercioby

It can be solved.It has an infinity of solutions.Here's how to do it without Maple...

Under the substitution

$$e^{-i \alpha}=x$$

the initial equation

$$\frac{e^{i\alpha}+e^{-i\alpha}}{2}+\left(\frac{e^{i\alpha}+e^{-i\alpha}}{2}\right)^{2}+\left(\frac{e^{i\alpha}+e^{-i\alpha}}{2}\right)^{4}=4$$

becomes a reducible 8-th order algebraic equation in "x".

Or simply make $\cos\alpha=x$ in the initial equation,find the 4 solutions in $\mathbb{C}$ and then solve for $\alpha$ in $\mathbb{C}$.

Daniel.

11. Apr 28, 2005

### Night Owl

Alright. The first part of your last post didn't quite make sense to me, but that's just because I haven't really learned a whole lot about how to do operations with complex numbers yet. I still need to finish deriving the taylor expansion (I might have the name wrong...the thing where you take the limit of the taylor series as n approaches infinity) for $f(x) = e^{i\alpha}$ before I'd really understand that. Setting $\cos\alpha = x$, however, does make sense to me.

Thanks for the feedback! I appreciate it.

Last edited: Apr 28, 2005
12. Apr 28, 2005

### Gunni

Does that equation have any real solutions? Or even complex ones?

We know that for all x that |cos x| <= 1, and therefore |cos^n x| <= 1. So based on that we get

cos x + cos^2 x + cos^4 x <= 1 + 1 + 1 = 3 < 4

so we must conclude that the original equation has no solutions.

13. May 1, 2005

### meister

It has no real solutions, Gunni, but in the complex plane |Cos x| can be greater than 1.

14. May 1, 2005

### Orion1

Identity Theft...

Does reducing such a high exponential equation into a lower expontial 'identity' produce better results?

$$\cos x + \cos^2 x + \cos^4 x = 4$$
$$\cos x(1 + \cos x + \cos^3 x) = 4$$
$$\cos x(1 + \cos^2 x) = 4 \sec x - 1$$
$$1 + \cos^2 x = 4 \sec^2 x - \sec x$$
$$\cos^2 x = 4 \sec^2 x - \sec x - 1$$
$$\boxed{\cos x = \pm \sqrt{4 \sec^2 x - \sec x - 1}}$$

And for the zeros?
$$\boxed{\cos x = \pm \sqrt{ \left( \sec x - \frac{1 - \sqrt{17}}{8} \right) \left( \sec x - \frac{1 + \sqrt{17}}{8} \right)}}$$

15. May 2, 2005

### DuncanM

This is a simple fourth-degree polynomial.

Just make the substitution cos(alpha) = x

Then the equation becomes

x^4 + x^2 + x - 4 = 0

Simply solve the quartic for x (i.e. - cos(alpha)).
For example, an online utility is posted here:

You should get two real solutions and two complex solutions.
Knowing cos(alpha), you can then determine alpha.

Note two of the results are greater than 1, which doesn't make sense for cosine. That implies they should be ignored, and only the complex results taken into consideration.

Last edited by a moderator: May 2, 2017
16. May 2, 2005

### saltydog

Well, I did and I would like to recommend that others go through the trouble of working out the complex solution by hand (if you could use the practice). Solve the quartic as Daniel suggested (I used Mathematica for that part), and then use the following relations:

$$Cos^{-1}(z)=\frac{\pi}{2}+iLn[iz+\sqrt{1-z^2}]$$

with:

$$Ln(z)=Ln(r)+i\theta$$

$$\theta=ArcTan[\frac{Im(z)}{Re(z)}]$$

$$r=\sqrt{Re(z)^2+Im(z)^2}$$

Anyway, I'll give the first one if x=Cos($\alpha$).

$$x_1=0.120841+1.60731i$$

Doing all the above I get:

$$\alpha_1=1.50701-1.25458i$$

Thanks, interesting problem.