Trigonometric Equation solving

[bnosam]

Homework Statement

cos 2x + cos 4x = cos x

0 degrees ≤ X ≤ 360 degrees

Homework Equations

Trig identities.

The Attempt at a Solution

Trig identities and equations are my weakness so I'm trying to review some to get better at them, so I guess this isn't technically homework. But it's out of my textbook.

Can someone give me a kick for the right start?

I tried:

2cos(\frac{(2x + 4x)}{2}) cos(\frac{(2x-4x)}{2})

2cos3x * cosx = cosx

2cos3x = 1

cos3x = 1/2

I'm not sure if this is right or where to go from here.

 

[Mark44]

Homework Statement

cos 2x + cos 4x = cos x

0 degrees ≤ X ≤ 360 degrees

As stated, it's not clear what you need to do. Are you supposed to solve an equation or prove an identity? From your work below, you are apparently solving an equation.

Homework Equations

Trig identities.

The Attempt at a Solution

Trig identities and equations are my weakness so I'm trying to review some to get better at them, so I guess this isn't technically homework. But it's out of my textbook.

Can someone give me a kick for the right start?

I tried:

2cos(\frac{(2x + 4x)}{2}) cos(\frac{(2x-4x)}{2})

How is this related to cos(2x) + cos(4x)? It could be that the above equals cos(2x) + cos(4x), but this is not an identity that I have committed to memory.

2cos3x * cosx = cosx

2cos3x = 1

Assuming for the moment that your work is correct, you should never do what you just did, since you lost a bunch of solutions. You pretty much never want to divide by a variable expression, because it's possible that expression might be zero for some value(s) of the variable.

Instead, bring everything over to one side, and then factor it.

cos3x = 1/2

I'm not sure if this is right or where to go from here.

Can you solve cos(b) = 1/2?

 

[CAF123]

2cos(\frac{(2x + 4x)}{2}) cos(\frac{(2x-4x)}{2})

2cos3x * cosx = cosx *

2cos3x = 1

cos3x = 1/2

I'm not sure if this is right or where to go from here.

On line *, where you 'divide out' by cosx, you are implicitly assuming cosx is not zero here, that is that x is not $\pi/2 + n\pi$. Do you think this is wise?

 

[bnosam]

As stated, it's not clear what you need to do. Are you supposed to solve an equation or prove an identity? From your work below, you are apparently solving an equation.
How is this related to cos(2x) + cos(4x)? It could be that the above equals cos(2x) + cos(4x), but this is not an identity that I have committed to memory.
Assuming for the moment that your work is correct, you should never do what you just did, since you lost a bunch of solutions. You pretty much never want to divide by a variable expression, because it's possible that expression might be zero for some value(s) of the variable.

Instead, bring everything over to one side, and then factor it.

Can you solve cos(b) = 1/2?

Trying to solve the equation to find the angle.

It is related to cos(2x) + cos(4x) because those are the identities for addition of cos.

cos u + cos b = 2cos((u+v)/2) * cos((u-v)/2)

So I guess what I did was wrong then? So I have no idea where to go, unfortunately.

I just find everything in trig so far to be confusing, I was just starting to understand using the identities and so on, but coming to this, just confuses me and I can't get around the mental block of it.

 

[Mark44]

Trying to solve the equation to find the angle.

It is related to cos(2x) + cos(4x) because those are the identities for addition of cos.

cos u + cos b = 2cos((u+v)/2) * cos((u-v)/2)

So I guess what I did was wrong then? So I have no idea where to go, unfortunately.

I didn't say that. The identity you used is fine, I believe. It's just not one that I've used a lot.

I just find everything in trig so far to be confusing, I was just starting to understand using the identities and so on, but coming to this, just confuses me and I can't get around the mental block of it.

You got to here:
2cos(3x)cos(x) = cos(x)

Bring everything over to one side and then factor what you have.

 

[bnosam]

I didn't say that. The identity you used is fine, I believe. It's just not one that I've used a lot.




You got to here:
2cos(3x)cos(x) = cos(x)

Bring everything over to one side and then factor what you have.

2cos(3x)cos(x) - cos(x) = 0

cos(x) ( 2cos(3x) - 1) = 0

so cos(x) = 0
x = 90 , x = 180


2cos(3x) - 1 = 0

2(cos(3x) = 1

cos(3x) = 1/2

IS this right so far?

 

[Mark44]

2cos(3x)cos(x) - cos(x) = 0

cos(x) ( 2cos(3x) - 1) = 0

so cos(x) = 0
x = 90 , x = 180


2cos(3x) - 1 = 0

2(cos(3x) = 1

cos(3x) = 1/2

IS this right so far?

Mostly. cos(x) = 0 ==> x = 90° but does not imply that x = 180°. cos(180°) = -1.

Everything else looks fine.

A question I asked earlier was this: Can you solve cos(b) = 1/2?

 

[bnosam]

Mostly. cos(x) = 0 ==> x = 90° but does not imply that x = 180°. cos(180°) = -1.

Everything else looks fine.

A question I asked earlier was this: Can you solve cos(b) = 1/2?

Oops, wasn't thinking when I put 180. I meant 270.

cos(b) = 1/2
b = 60 degrees
cos 3x = 1/2

cos x = 60

3x = 60

x = 20So it checks correct when I sub it back in the equation. Did I get everything?

 

[Mark44]

Oops, wasn't thinking when I put 180. I meant 270.

cos(b) = 1/2
b = 60 degrees

Actually, b = 60° + k * 360°, where k is an integer

cos 3x = 1/2

[STRIKE]cos x = 60[/STRIKE]
3x = 60

x = 20

3x = 60° + k * 360°, where k is an integer
So x = ?

You want values of x such that 0 ≤ x ≤ 360°

So it checks correct when I sub it back in the equation. Did I get everything?

See above.

 

[bnosam]

Actually, b = 60° + k * 360°, where k is an integer
3x = 60° + k * 360°, where k is an integer
So x = ?

You want values of x such that 0 ≤ x ≤ 360°


See above.

3x = 60° + k * 360°, where k is an integer
So x = ?

would it be 20° + k * 360°?

 

[Mark44]

3x = 60° + k * 360°, where k is an integer
So x = ?

would it be 20° + k * 360°?

No.
You're multiplying both sides of the equation 3x = 60° + k * 360°, but you forgot to distribute the 1/3 to both terms.

 

[bnosam]

No.
You're multiplying both sides of the equation 3x = 60° + k * 360°, but you forgot to distribute the 1/3 to both terms.

x = 20 + (360k)/3

 

[CompuChip]

x = 20 + (360k)/3

Yep. And (360k)/3 = ...k ?

 

[bnosam]

Yep. And (360k)/3 = ...k ?

120k

 

[Mark44]

Yes, but you're not quite finished.
So x = 20° + k * 120°, for k an integer.

In the problem statement, 0° ≤ x ≤ 360°, so what values of k give x values in this range?

 

[Infrared]

Yes, but you're not quite finished.
So x = 20° + k * 120°, for k an integer.

I think you will be missing solutions. How will you get x=100°,220°, or 340° from this? cos(3x)=1/2 does not mean that 3x=60°+360°k for integer k.

 

[bnosam]

yes, but you're not quite finished.
So x = 20° + k * 120°, for k an integer.

In the problem statement, 0° ≤ x ≤ 360°, so what values of k give x values in this range?

20 + 1 * 120 = 140

20 + 2 * 120 = 260

 

[Mandelbroth]

20 + 1 * 120 = 140

20 + 2 * 120 = 260

0 is an integer too!

So, what is our complete solution set? Id est,

$$\cos{2x}+\cos{4x}=\cos{x} \\ (0^o \leq x \leq 360^o) \\ \forall x\in\{?\}$$

 

[bnosam]

0 is an integer too!

So, what is our complete solution set? Id est,

$$\cos{2x}+\cos{4x}=\cos{x} \\ (0^o \leq x \leq 360^o) \\ \forall x\in\{?\}$$

Our answers should be: 90°, 270°, 20°, 140°, 260°

 

[Infrared]

Reread my post #16. There are other solutions you can get from saying 3x=300+360k. You will see that x=100°,220°, and 340° will also satisfy your equation.

 

[bnosam]

Reread my post #16. There are other solutions you can get from saying 3x=300+360k. You will see that x=100°,220°, and 340° will also satisfy your equation.

Ok thank you. I understand this now.

I have another question that I'm not quite sure with.

sin3x = sinx

I tried

sin(2x + x) = sinx

sin2xcosx + cos2xsinx = sinx

sin2xcosx + cos2xsinx - sinx = 0

sin2xcosx + sinx(cos2x - 1) = 0

sin x = 0
x = 0, 180, 360

cos 2x -1 = 0

cos 2x = 1

2x = 0 + 360k
x = 180k

x = 0,180, 360 also.I'm a little stuck so far now

 

[CAF123]

Ok thank you. I understand this now.

I have another question that I'm not quite sure with.

sin3x = sinx

I tried

sin(2x + x) = sinx

sin2xcosx + cos2xsinx = sinx

sin2xcosx + cos2xsinx - sinx = 0

From here, one thing to try would be substitute an alternate expression for cos2x such that the LHS is entirely in terms of sinx.

sin2xcosx + sinx(cos2x - 1) = 0

sin x = 0
x = 0, 180, 360

cos 2x -1 = 0

cos 2x = 1

2x = 0 + 360k
x = 180k

x = 0, 180, 360 also.


I'm a little stuck so far now

$a + b = 0$ does not mean that $b$ need be zero.

 

[bnosam]

From here, one thing to try would be substitute an alternate expression for cos2x such that the LHS is entirely in terms of sinx.
$a + b = 0$ does not mean that $b$ need be zero.

Let me show more work then and see if this is what you mean:sin(2x + x) = sinx
sin2x cosx + cos2x sinx = sinx

sin2x cosx + sin² (x) sinx - sinx = 0

2sinx cos²(x) + sin³(x) - sinx = 0

sinx(2cos²(x) + sin²(x) - 1) = 0

sin x = 0
(2cos² (x) + sin²(x) - 1) = 0
2sin²(x) + sin²(x) -1 = 0
3sin²(x) -1 = 0

sin²(x) = 1/3

sinx = ± \frac{√3}{3}

Correct so far?

 

[Mark44]

Let me show more work then and see if this is what you mean:


sin(2x + x) = sinx
sin2x cosx + cos2x sinx = sinx

sin2x cosx + sin² (x) sinx - sinx = 0

2sinx cos²(x) + sin³(x) - sinx = 0

sinx(2cos²(x) + sin²(x) - 1) = 0

sin x = 0



(2cos² (x) + sin²(x) - 1) = 0
2sin²(x) + sin²(x) = 0

How did you get the equation above?

3sin²(x) -1 = 0

sin²(x) = 1/3

sinx = ± \frac{√3}{3}

Correct so far?

 

[bnosam]

How did you get the equation above?

I'm not quite sure waht you mean, did you mean the sin ( 2x + x)?

If so:
sin(U + V) = sin U cos V + cos U sin V

 

[Mark44]

No, I meant the equation directly above my question: 2sin²(x) + sin²(x) = 0.

How does this follow from 2cos²(x) + sin²(x) - 1 = 0?

 

[bnosam]

No, I meant the equation directly above my question: 2sin²(x) + sin²(x) = 0.

How does this follow from 2cos²(x) + sin²(x) - 1 = 0?

It was a typo, sorry. I corrected it on the post above

 

[Mark44]

"Corrected" is the wrong word. You just made it different, but no correct.

How did you go from

2cos²(x) + sin²(x) - 1 = 0
to
2sin²(x) + sin²(x) - 1 = 0
?

There is no identity that says that cos²(x) = sin²(x).

 

[bnosam]

"Corrected" is the wrong word. You just made it different, but no correct.

How did you go from

2cos²(x) + sin²(x) - 1 = 0
to
2sin²(x) + sin²(x) - 1 = 0
?

There is no identity that says that cos²(x) = sin²(x).

My book has cos 2x = cos²(x) = sin²(x) as an identity in the list of identities...

It must be a typo in the book, the identity must be cos 2x = cos²x - sin²x.

So confusing, no wonder my answer was wrong!

 

[Mark44]

My book has cos 2x = cos²(x) = sin²(x) as an identity in the list of identities...

Either you're not reading your book correctly, or the book has a typo in a very important identity.
It should read cos 2x = cos²(x) -[/color] sin²(x)

That's "minus" NOT "equals".

 

[bnosam]

Either you're not reading your book correctly, or the book has a typo in a very important identity.
It should read cos 2x = cos²(x) -[/color] sin²(x)

That's "minus" NOT "equals".

Yeah, it was a typo on their part I guess. I'll redo the question.

continuing from after sinx = 02cos²x + sin²x -1 = 0

2(1-2sin²x) + sin²x - 1 = 0
1 - 3sin²x = 0
3sin²x = 1

sin²x = 1/3

I get the same answer? I must be doing something wrong.

 

[Mark44]

Yeah, it was a typo on their part I guess.

If it was my book, I would replace that '=' with '-' using a heavy pen.

I'll redo the question.

continuing from after sinx = 0


2cos²x + sin²x -1 = 0

2(1-2sin²x) + sin²x - 1 = 0

The above is wrong. You replaced cos²(x) with 1 - 2sin²(x).

The identity you should be thinking of is sin²(x) + cos²(x) = 1. If you solve for cos²(x) in this identity, what do you get?

1 - 3sin²x = 0
3sin²x = 1

sin²x = 1/3

I get the same answer? I must be doing something wrong.

 

[bnosam]

If it was my book, I would replace that '=' with '-' using a heavy pen.
The above is wrong. You replaced cos²(x) with 1 - 2sin²(x).

The identity you should be thinking of is sin²(x) + cos²(x) = 1. If you solve for cos²(x) in this identity, what do you get?

I need to pay more attention reading the identities, sorry.

sin3x = sinx
sin(2x + x) = sin2x cosx + sinx cos2x
(2sinxcosx)cosx + sinx(2cos²x - 1) = sinx
2sinx cos²x + 2cos²x sinx - 2 sinx
4sinxcos²x - 2sinx = 0

sinx(2cos²x - 1) = 0

sinx = 0


2cos²x -1 = 0
cos²x = 1/2
cos x = ± √2/2

 

[Mark44]

I need to pay more attention reading the identities, sorry.

sin3x = sinx
sin(2x + x) = sin2x cosx + sinx cos2x Put this line to the side. It just explains which identity you're using.[/color]
(2sinxcosx)cosx + sinx(2cos²x - 1) = sinx
2sinx cos²x + 2cos²x sinx - 2 sinx = 0[/color]
4sinxcos²x - 2sinx = 0

sinx(2cos²x - 1) = 0

sinx = 0

OR[/color]
2cos²x -1 = 0
cos²x = 1/2
cos x = ± √2/2

This looks good, now, but you're not done yet.

If sin(x) = 0, x = ? With no restrictions, there are an infinite number of values.
Or if cos(x) = ± √2/2, x = ? As above, with no restrictions given, there are an infinite number of values.

 

[bnosam]

This looks good, now, but you're not done yet.

If sin(x) = 0, x = ? With no restrictions, there are an infinite number of values.
Or if cos(x) = ± √2/2, x = ? As above, with no restrictions given, there are an infinite number of values.

The interval of values to solve for is 0° ≤ x ≤ 360°.

so x in sin x= 0 can be equal to 0°, 180° and 360° in that interval


x in cos x = ±√2/2

so 45°, 135°, 225°, 315° are valid

 

[Mark44]

Looks good.

A good thing to do is to check to confirm that sin(3x) = sin(x) for each of the seven values you have.

 

[epenguin]

Haven't followed this but you said trig identities. The trig identity to me most obviously relevant to problem #1 is the one for cos2x in terms of cos x, have you tried that?

(I think that gives you one of your (families of) solutions, and there is only one other real... if it's 20° as was said, that's' nothing very special or interesting to spend your time on IMHO).