Pre Calc take home test!

[Diego14]

alright I have to hand this in tomorrow, I'll be up all night writing a term paper. This is from our retake exam 3, It was take home because the average of the class failed. I'm not asking for you to give me the answer, I'm genuinely lost and looking for a tip. I'm in pre-calculus, thanks.
5) Determine the solutions of the equation sec(3x)=cos(3x) that like in the interval 0degrees greater than or equal x greater than or equal to 360degrees

 

[Diego14]

how do you get the math symbols on this forum.?

 

[Diego14]

as in help i mean you know what part that would be in either chapter 5 Trig Functions of Real Numbers or 7 Analytic Trig i thought i had an idea but until i understand how to insert symbols...Mods.

 

[DrMike]

alright I have to hand this in tomorrow, I'll be up all night writing a term paper. This is from our retake exam 3, It was take home because the average of the class failed. I'm not asking for you to give me the answer, I'm genuinely lost and looking for a tip. I'm in pre-calculus, thanks.
5) Determine the solutions of the equation sec(3x)=cos(3x) that like in the interval 0degrees greater than or equal x greater than or equal to 360degrees

Hint : sec(3x) = 1/cos(3x).

What have you tried so far?

 

[Diego14]

oh as for that i set the problem up like that, i dont want the answer just dont know what to do. I set it up as
1 over cos(3x)=cos(2x+1x) and used the formula for 2cos. But idk if that's right. I really wanna know what the 0-360 degrees mean, in relations to the problem. I know the rotation on the graph, (x,y) (cos,sin)

 

[BigGlenntheHeavy]

\(\displaystyle sec(3x) \ = \ cos(3x)\)

\(\displaystyle \frac{1}{cos(3x)} \ = \ cos(3x)\)

\(\displaystyle cos^{2}(3x) \ = \ 1\)

\(\displaystyle cos(3x) \ = \ 1\)

\(\displaystyle 3x \ = \ arccos(1)\)

\(\displaystyle 3x \ = \ 0\)

\(\displaystyle x \ = \ 0\)

 

[Diego14]

\(\displaystyle sec(3x) \ = \ cos(3x)\)

\(\displaystyle \frac{1}{cos(3x)} \ = \ cos(3x)\)

\(\displaystyle cos^{2}(3x) \ = \ 1\)

\(\displaystyle cos(3x) \ = \ 1\)

\(\displaystyle 3x \ = \ arccos(1)\)

\(\displaystyle 3x \ = \ 0\)

\(\displaystyle x \ = \ 0\)

thanks?! id really like to know how you did it...I dont understand what the ? asks me to do, that's why I came here. Can you help?

 

[wjm11]

5) Determine the solutions of the equation sec(3x)=cos(3x) that like in the interval 0degrees greater than or equal x greater than or equal to 360degrees

Diego14,

BigGlenn has given you *one* of the possible answers. Your job is to figure out if there are others. For example, if you plug Glenn’s answer into the original problem, you get

sec(3x)=cos(3x)
sec(3(0))=cos(3(0))
sec(0) = cos (0)
1 = 1

There may be other values of x that will work, too. You are supposed to find all the values of x between 0 and 360 that will make the equation true.

To see what is going on graphically, plot these two equations on the same graph:

y = sec(3x)
y = cos(3x)

See where the two equations intersect each other (between x = 0 and x = 360). Those intersections are your solutions. Make sense? (Hint: the other solutions also occur when sec(3x) and cos(3x) both equal 1.)

 

[chrisr]

Hi Diego,

wjm11's response and the earlier posts in answer to your question
are extremely good in terms of helping you improve your skill very quickly.

If x must lie between 0 and 360 degrees,
(have a look at how you worded that again)
then what is the range of angles 3x lie between ?
given that 370 degrees equals 10 degrees after one revolution etc

 

[Diego14]

so your saying 370 goes through ten, which is the same as one full revolution? As in the solution thats fine with what 1=1 if you scroll up. But for all solutions how would i find it. I would just plug in y = sec(3x)and y = cos(3x) to find my answers??

 

[chrisr]

Hi Diego,

Yes, if you had a graph with both Sec3x and Cos3x on it,
you would see all the points of intersection where Sec3x = Cos3x.

I made an Adobe .pdf file showing them and tried to upload it
for you but got an error ".pdf not supported".
Sorry about that.

Mathematically you can proceed in a few different ways.
Since Sec3x is 1/(Cos3x) then {Cos3x}^2 = 1 so Cos3x = +-1.
You have to take the negative square root also because in the circle of radius 1,
centred at the origin (0,0), Cos(angle) is the x co-ordinate of a point on
the circle circumference.

This is 1 for 0, 360, 720 degrees etc and -1 for 180, 540 degrees etc

Unfortunately, I couldn't show you the graph, but Cos3x has a period of 120 degrees as does Sec3x.

Since Cos3x = Sec3x for 3x = 0, 180, 360, and so on,
then Cos3x = Sec3x for x = 0, 60, 120, 180, 240, 300 and 360 degrees if x can range from 0 to 360 degrees

Another way to look at it is as follows.
Cos(0) = 1 so 1/{Cos(0)} = 1 and Cos(180) = -1 so 1 over that is also -1.
For any other angle, Cos(angle) is between 0 and 1 so 1 over that is >1
or between 0 and -1 so 1 over that is between -1 and -infinity, meaning they cannot be equal for
angles off the x axis.
Therefore they can be equal only at 0, 180, 360, 540 etc

It's important to realise that the x co-ordinate of a point on the circumference of
the unit circle centred at (0,0) is Cos(angle).
That's how the calculator uses the function.

I'm assuming you meant "x is greater than or equal to zero and less than or equal to 360 degrees".