Solving 1 + 2cos(2x+ π/3) for Zeroes: Step-by-Step Guide

[jasper10]

Homework Statement

Find the zeroes of

f(x) = 1 + 2cos(2x+ π/3)

Range: (0 ; π)

The Attempt at a Solution

1 + 2cos(2x + π/3) = 0

hence, cos(2x + π/3) = -1/2

hence, 2x + π/3 = 2π/3

and, x = π/6 which is 30 degrees

can someone help me find the other 0 ?

I don't know how to!

Thanks a lot!

(the answer is x=π/2 but i don't know how to get there!)

 

[Char. Limit]

First, I don't quite know how you went from your first step to your second, even though it does turn out to be correct. A bit of explanation would be most appreciated.

Second, have you tried using the sum-of-angles formula to expand your cosine out? It might make it easier, I think...

 

[jasper10]

Ok Char. Limit,

I edited my answer and showed the transition step.

What do you mean by "the sum of angles formula" ?

 

[Sir Beaver]

Well, if one looks at the unit circle, one can see that 2*pi / 3 is one place where
cos(y) = -1/2. However, there are more places on the unit circle in which
cos(y) = -1/2. Finally, one can also add 2*pi*n to the result, and end up at
the same place.

 

[Char. Limit]

Well, if one looks at the unit circle, one can see that 2*pi / 3 is one place where
cos(y) = -1/2. However, there are more places on the unit circle in which
cos(y) = -1/2. Finally, one can also add 2*pi*n to the result, and end up at
the same place.

But he restricted x to be between zero and pi...

The sum-of-angles formula is as follows:

cos(x+y)=cos(x)cos(y)-sin(x)sin(y)

I have no idea how to prove it, but it is a theorem, I believe.

 

[Sir Beaver]

Yes, but still, one can manually inspect the formulas to see when
<br /> 2x + \pi /3 = 2\pi /3 +2\pi n<br />
, and after that see if x is inside the interval between 0 and pi.

 

[jasper10]

However, there are more places on the unit circle in which
cos(y) = -1/2.

Yes, my question ultimately is

How does one determine these other places where cos(y) = -1/2 ?

 

[Sir Beaver]

If you are familiar with the unit circle, you have that the x-coordinate on the circle
is cos(x), while the y-coordinate is sin(x). For a picture, please see
http://en.wikipedia.org/wiki/Unit_circle
In the picture, you can see that cos(x) = -1/2 means: at which place at the unit circle
is the x-coordinate equal to -1/2? And there are two places that can happen at in the circle. Does this help?

 

[Yousef]

How does one determine these other places where cos(y) = -1/2 ?

Well, wouldn't you just have to add pi/4 for the next zero and then 3pi/4 and keep going until you're beyond your range in this case? Since the function is pretty much a squashed together cos curve (twice as many zeros as a normal one) that is half it's amplitude up the y-axis, it's amplitude being 2, that is out of phase of your standard cos curve by pi/3. I'm not sure I'm visualising it right, but I guess you could always sketch it up on some spare sheet.

 

[Mark44]

Correction: cos(y) = -1/2 for y = 2pi/3 or for y = 4pi/3.
Replace y by 2x + pi/3 and you should get two values of x in (0, pi).

 

[Mark44]

Second, have you tried using the sum-of-angles formula to expand your cosine out? It might make it easier, I think...

It actually makes it much harder.

 

[Char. Limit]

It actually makes it much harder.

I stand corrected then.

 

[jasper10]

cos(y) = -1/2 for y = 4pi/3 or for y = 5pi/3.

How do you know this?

 

[Cyosis]

Mark gave you the values where the cosine is -1/2 in the third quadrant and 1/2 in the fourth quadrant. You're interested in the values where cosine is -1/2, not 1/2.

The first thing you should do is draw a unit circle. Then sketch the places where the cosine is equal to -1/2 and find the appropriate angles. In which two quadrants does your solution lay?

 

[jasper10]

The first thing you should do is draw a unit circle. Then sketch the places where the cosine is equal to -1/2 and find the appropriate angles. In which two quadrants does your solution lay?

How does one find the place where cosine is equal to -1/2?

and as you said, the solution is in the third quadrant and the first solution (Pi/6) is in the first.

 

[Mark44]

Mark gave you the values where the cosine is -1/2 in the third quadrant and 1/2 in the fourth quadrant. You're interested in the values where cosine is -1/2, not 1/2.

Due to a mental lapse, one of the values I gave was incorrect. The values of y should have been in the second and third quadrants, but not in the fourth. I have corrected my earlier response.

The first thing you should do is draw a unit circle. Then sketch the places where the cosine is equal to -1/2 and find the appropriate angles. In which two quadrants does your solution lay?

 

[Cyosis]

No, \cos \pi/6=\frac{1}{2}\sqrt{3}. Secondly I am not asking you to draw the position where cos is -1/2 in the unit circle exactly. The use in the unit circle is that you can see in which quadrant the solutions lay. When you draw a line radially outwards from the origin you can form a triangle. Which side of that triangle corresponds to the cosine and which one to the sine? To answer your question you want to know where that side of the triangle is -1/2.

 

[Mark44]

How does one find the place where cosine is equal to -1/2?

There are just a handful of angles that you are expected to know exactly: 0, pi/6, pi/4, pi/3, pi/2, plus their complements and supplements. If you plot these angles in the unit circle, you can use symmetry to show that cos(y) = 1/2 for y = pi/3 and y = 5pi/3, and that cos(y) = -1/2 for y = 2pi/3 or y = 4pi/3.

and as you said, the solution is in the third quadrant and the first solution (Pi/6) is in the first.

 

[jasper10]

Ok! thanks, i kind of get it now!