Solve for all angles x: cos(2x) + cos(x) = 0, where 0<x<2pi

[Ascendant0]

Homework Statement

The title says it all, solve for all angles that can be used in the equation

Relevant Equations

If I knew, I wouldn't be asking

I'm not sure how to go about solving this mathematically? In just using what seems obvious, I know the angle pi would work, because pi = -1, and 2pi = 1. However, as far as manipulating the equations in a way where it can solve itself without me having to look at a chart where cos for both x and 2x will cancel each other out, I don't know what kind of formula(s) would help me solve it without referring to charts?

 

[renormalize]

I'm not sure how to go about solving this mathematically?

How about making the substitution $\cos\left(nx\right)\rightarrow\frac{1}{2}\left(e^{inx}+e^{-inx}\right)$ and then seeing if you can factor the resulting sum of exponentials into something you can solve for $x$?

 

[Hill]

I don't know what kind of formula(s) would help

A formula that connects cos(2x) to cos(x).

 

[Ascendant0]

How about making the substitution $\cos\left(nx\right)\rightarrow\frac{1}{2}\left(e^{inx}+e^{-inx}\right)$ and then seeing if you can factor the resulting sum of exponentials into something you can solve for $x$?

Thank you for the reply. Unfortunately, that is way outside of the mathematics I'm working on currently. If you know of any resources online that would cover working on it in that way, or additional steps I'd appreciate it. I'd love to better understand where you're going with that. Unfortunately, it's definitely not the way they're expecting us to solve it at this point in Calc though.

The more I'm looking through this chapter, the more I'm thinking they are just expecting us to look at the reference charts with the list of common angles to solve this (see which ones at 2x cancel x). Not really sure of any other way we could do it at this point of calc.

 

[PeroK]

Using $\cos(2x) = 2\cos^2x -1$ leads to a quadratic equation in $\cos x$.

 

[Ascendant0]

A formula that connects cos(2x) to cos(x).

Thank you. So far, the only formula we've been shown that does that (at least from the textbook, Calculus by Thomas) would be the double angle formula:

cos(2x) = cos^2(x) - sin^2(x)

But, I don't see how that helps, as then, it adds sin into the equation as well.

So far, what we have to work with that could potentially be relevant is the trig identities and double angle formulas. I don't see anything else in this chapter (or prior ones) that would be relevant to how they expect us to solve it at this point.

 

[PeroK]

Thank you. So far, the only formula we've been shown that does that (at least from the textbook, Calculus by Thomas) would be the double angle formula:

cos(2x) = cos^2(x) - sin^2(x)

But, I don't see how that helps, as then, it adds sin into the equation as well.

So far, what we have to work with that could potentially be relevant is the trig identities and double angle formulas. I don't see anything else in this chapter (or prior ones) that would be relevant to how they expect us to solve it at this point.

You give up too easily!

 

[renormalize]

cos(2x) = cos^2(x) - sin^2(x)

But, I don't see how that helps, as then, it adds sin into the equation as well.

Use $\text{sin}^2(x)=1-\text{cos}^2(x)$.

 

[Hill]

Thank you. So far, the only formula we've been shown that does that (at least from the textbook, Calculus by Thomas) would be the double angle formula:

cos(2x) = cos^2(x) - sin^2(x)

But, I don't see how that helps, as then, it adds sin into the equation as well.

So far, what we have to work with that could potentially be relevant is the trig identities and double angle formulas. I don't see anything else in this chapter (or prior ones) that would be relevant to how they expect us to solve it at this point.

This is all you need. It is a four-step process:
1. Eliminate 2x in favor of x.
2. Eliminate sin in favor of cos.
3. Solve for cos(x).
4. Solve for x.

 

[WWGD]

Thank you. So far, the only formula we've been shown that does that (at least from the textbook, Calculus by Thomas) would be the double angle formula:

cos(2x) = cos^2(x) - sin^2(x)

But, I don't see how that helps, as then, it adds sin into the equation as well.

So far, what we have to work with that could potentially be relevant is the trig identities and double angle formulas. I don't see anything else in this chapter (or prior ones) that would be relevant to how they expect us to solve it at this point.

You can turn it into a quadratic on $cosx$, with $sin^2x$ the constant term, and use the quadratic equation.

 

[Mark44]

So far, the only formula we've been shown that does that (at least from the textbook, Calculus by Thomas) would be the double angle formula:

cos(2x) = cos^2(x) - sin^2(x)

I would think you would also have learned this one, one of the most basic trig identities:
$\cos^2(x) + \sin^2(x) = 1$

This would allow you to replace the $-\sin^2(x)$ term in the double angle formula with $\cos^2(x) - 1$ to get $\cos(2x) = \cos^2(x) + \cos^2(x) - 1$.

Use this in your original equation to get a quadratic in $\cos(x)$.

 

[Ascendant0]

Thank you for all the help, all of you! Now it makes more sense, but I'm still thrown off on one of three of the possible solutions in the solutions book.

I did try this suggested method earlier, but when I solved, the numbers were different than the solution answers, and so thought I did it wrong.

Once I saw all of you suggesting it, I tried it again, and I realized that the answers I got, x = 1/2 and x = -1 were correct for the x values, but then I needed to use arccos afterwards to solve x for cos. Then, the answers were the same as the book, other than the book using pi in the answers (but same values when calculated out).

The last thing that's throwing me off though is that I got the 1/2 and -1 answers, which matches pi/3 and pi from in the book, but the book also has 5pi/3 as an additional answer, which I did not get from using the quadratic formula. I've been racking my brain as to why solving for x didn't give that answer as well to no avail. What am I missing in that answer that's keeping me from getting the 5pi/3 answer as well?

 

[Hill]

Thank you for all the help, all of you! Now it makes more sense, but I'm still thrown off on one of three of the possible solutions in the solutions book.

I did try this suggested method earlier, but when I solved, the numbers were different than the solution answers, and so thought I did it wrong.

Once I saw all of you suggesting it, I tried it again, and I realized that the answers I got, x = 1/2 and x = -1 were correct for the x values, but then I needed to use arccos afterwards to solve x for cos. Then, the answers were the same as the book, other than the book using pi in the answers (but same values when calculated out).

The last thing that's throwing me off though is that I got the 1/2 and -1 answers, which matches pi/3 and pi from in the book, but the book also has 5pi/3 as an additional answer, which I did not get from using the quadratic formula. I've been racking my brain as to why solving for x didn't give that answer as well to no avail. What am I missing in that answer that's keeping me from getting the 5pi/3 answer as well?

There are two values of x in the given range that make cos(x)=1/2.

 

[PeroK]

.If the question asked for values $0<x<4\pi$ then you would have more solutions. And, if it had asked for all solutions, there are an infinite number of them.

On a calculator, $\arccos$ will only give you one value in the range $[0,\pi]$.

Note that you could have drawn a graph of the two functions to see how many solutions there are.

 

[pasmith]

You can turn it into a quadratic on $cosx$, with $sin^2x$ the constant term, and use the quadratic equation.

Except that $\sin^2 x$ is not a constant independent of $\cos x$, but is equal to $1 - \cos^2 x$.

 

[WWGD]

Except that $\sin^2 x$ is not a constant independent of $\cos x$, but is equal to $1 - \cos^2 x$.

That doesn't matter , $\frac {-1 \pm \sqrt {1+4sin^2(x)}}{2}$ is a zero of $cos^2x+cosx -sin^2x$.

 

[PeroK]

That doesn't matter , $\frac {-1 \pm \sqrt {1+4sin^2(x)}}{2}$ is a zero of $cos^2x+cosx -sin^2x$.

That's certainly a unique piece of mathematics!

 

[vela]

So far, the only formula we've been shown that does that (at least from the textbook, Calculus by Thomas) would be the double angle formula

There seems to be a mismatch in what you think you need to know and what the book's authors expect you to know. Presumably, if you're learning calculus, you have already mastered basic algebra and trig, and it's expected that you can actually recall that stuff on your own.

 

[vela]

You can turn it into a quadratic on $\cos x$, with $\sin^2 x$ the constant term, and use the quadratic equation.

While you can certainly do that, why would you want to?

 

[gmax137]

Note that you could have drawn a graph of the two functions to see how many solutions there are.

That's the first thing I did, it took less than a minute in excel.

 

[WWGD]

While you can certainly do that, why would you want to?

It was part of the prelim/bull session/ brainstorming session for the problem.

 

[Ascendant0]

There seems to be a mismatch in what you think you need to know and what the book's authors expect you to know. Presumably, if you're learning calculus, you have already mastered basic algebra and trig, and it's expected that you can actually recall that stuff on your own.

Thanks, and I'm trying to. Been years since I had been in college and done any of this. There's a ton to brush up on, so trying to brush up on my algebra, trig, and calc all at once

 

[Mark44]

There's a ton to brush up on, so trying to brush up on my algebra, trig, and calc all at once

But that's what you need to do. The good thing is that if you've been through it before (and were fairly capable at it), it's easier the second time through.

 

[Ascendant0]

But that's what you need to do. The good thing is that if you've been through it before (and were fairly capable at it), it's easier the second time through.

Thanks for the words of encouragement. They're greatly appreciated and needed at this point.

Back when I was attending college, I was more than capable. I finished my AA with a solid 4.0. I was so far ahead of classmates that I was an "unofficial" tutor for Physics I & II (the college didn't have the budget for a TA for undergrad, but the professor was also the chair of the Natural Science Department, pressed for time, and so I did it simply because I like helping others).

I had to stop college at the end of my 3rd year due to a family crisis. To add, I'm very self-critical, so even then, I would get frustrated with myself if I struggled to find the answer to a problem. If I didn't get every homework question right, and 100 on every exam, I would be extremely frustrated with myself. Always been very self-critical.

Trig was always my weak spot back then. I know it took a while for me to fully wrap my brain around some of it. The calculus involving derivatives, integrals, stepping it up into 3D (and 4D with time) in Calc III, no problem for me at all. But with trig, there were some things that were challenging for me. This time around, I know there is some of it that I grasp better this time around, but still a TON to catch up on.

Oh, and proofs. I was NEVER good at proofs, especially trig ones. I know I need to brush up on it a lot. I'm using both the Thomas Calc book and a trig and algebra book specifically designed to brush up on what you need for Calc ("Just in Time Algebra and Trigonometry for Calculus").

Sorry for the lengthy post. Just a bit frustrated and thoughts are racing. I have at least six courses I need to refresh on before I start college, and ideally looking to go in fall. So, six courses, six months, while I'm working full-time and still trying to make some time for family (we have children). It's do-able, but I really have to put this time in to have things committed to memory.

I really appreciate everyone's help. I definitely need it at times, lol.

 

[willem2]

Actually, this can be done without quadratic equations. If you get your equation in the form: cos(a) = cos(b), you can conclude a = b+2k pi or a = -b +2kpi. no double angle formula's or quadratic equations required.

 

[Hill]

Actually, this can be done without quadratic equations. If you get your equation in the form: cos(a) = cos(b), you can conclude a = b+2k pi or a = -b +2kpi. no double angle formula's or quadratic equations required.

This would give two answers, but the problem has three.

P.S. Sorry, I've made a mistake. I've missed, that one of these two answers allows for two different values of k.