Question

Find Solutions in interval "[0,2pi)"; cos^2 x + 2cosx + 1 = 0

Answer 1

I don't so much need the answer, just how to get it. I don't fully understand the concept of intervals and how to find it by applying this "[0,2pi)".

Answer 2

Trig functions are periodic, meaning that they will go on forever repeating the same kinds of values over and over. When they give you an interval, they are saying "just look in this neighborhood". When they give you 0 to 2pi, it is kind of nice because that is where you are used to looking.

Answer 3

Where those lines are crossing the x-axis on his graph...would those count as intervals?

Answer 4

They can, but the interval defined by your problem says to look from 0 (the y-axis) over to 2pi (the end of one cycle).

Answer 5

Ok, thank you. =) I have a poor understanding of the unit circle in general so it's nice to learn SOMETHING about it

Answer 6

They limited your interval (from being infinity) , because that would have given an infinite number of answers. They just wanted the standard answer that you would get if you looked "in the neigborhood" of where the function started.

Answer 7

You're welcome.

Answer 8

I hate to bother you further about the same concept, but I've never seen an equation of this caliber. 7tan^3 x - 21tan x = 0. Same coordinants; what can you get from that?

Answer 9

I would start by factoring out 7 tan x. Then set each factor to zero.

Answer 10

Alrighty

Answer 11

._. I'm trying to factor it out replacing tan with "a" but I'm having a bit of trouble. Do you have any hints or tips for me?

Answer 12

my first step looked like this\[7\tan x(\tan ^{2}x-3)=0\]

Answer 13

These kinds of problems are where algebra and trig collide. You look at whole trig functions as variables in a sense. Once we factor out... set each factor to 0.\[7\tan x=0\]and\[\tan ^{2}x-3=0\]

Answer 14

I only use a substitution if my factoring gets really hairy... like if it takes the form \[ax ^{2}+bx+c=0\]

Answer 15

How would I solve for x with "tan" in the equation?

Answer 16

inverse tangent... I had a teacher force me to learn my unit circle including tangents... so I think of these as "nice" answers. \[\tan ^{2}x=3\]\[\tan x=\pm \sqrt{3}\]\[\tan^{-1} (\tan x)=\tan^{-1} (\pm \sqrt{3})\]\[x=\pm \frac{\pi}{3}\]

Answer 17

cos^2 x + 2cosx +1=(cosz+1)^2=0
then
cosx=-1
ans=[pi,3pi,5pi,...,(2n+1)pi
give a medal plz

Answer 18

lol force? And I know 4pi/3 has to be an interval...would just "pi" we one as well? How can we test this? (I have a lot of questions about this, it's ok if you leave)

Answer 19

Ah... -pi/3 wasn't in the interval was it.... from 0 to 2pi.

Answer 20

You've got it... 4pi/3 is the other one... perfect! And I think that pi would be one as well as zero. You are getting the hang of this interval thing!

Answer 21

You could plug these numbers into your original equation to verify... just plug them in for x and see if you get zero.

Answer 22

=p Is that all the possible intervals? I can't think of anymore

Answer 23

My pc is doing strange things... forgive me if I lag. Remember that interval is not the answer it is simply the place where you are looking for answers.

Answer 24

Given the interval from zero (inclusive) to 2pi (non-inclusive), your answers should be 0, pi/3, pi, and 4pi/3... the next one would be 2pi, but that would be outside of the interval asked about.

Answer 25

ooh I think there are some hiding... remember when I said -pi/3... even though it is not in your given interval... it will produce solutions at 2pi/3 and 7pi/3.... I need to get my calculator. :)

Answer 26

Hmm...im not sure about 2pi/3, but wouldnt 7pi/3 be outside our [0,2pi) boundaries?

Answer 27

you are correct... brain fart... I meant 5pi/3

Answer 28

ok... so here is a way to gain a little insight.. do you have a graphing calculator?

Answer 29

No, but I can try to find an online one =o

Answer 30

I will give my final answer now :p {0, pi/3, 2pi/3, pi, 4pi/3, 5pi/3} What do you know... every pi/3.

Answer 31

Graph your original equation without the equals zero. Every x-axis crossing between [0, 2pi) is a solution.

Answer 32

And that's all within [0,2pi)? The > 2 digits are throwing me off

Answer 33

my answer is true
in your domain just pi can be a answer

Answer 34

fractions can make things look a little strange... if we are looking at pi/3's then 6pi/3 is 2pi and 5pi/3 is therefore the largest pi/3 we could use.

Answer 35

mahdi... we are on a different problem from the original posting

Answer 36

This one was a little rough... every time I thought I had it, there was more to it. Looking back... I would say that our original three answers (without considering interval) were -pi/3, 0, and pi/3. Due to the fact that a tangent has a period of pi, these solutions repeat every pi on the interval. If there was not a given interval... the solution would occur every pi/3 or generally x=npi/3 where n is an integer. The values of this that fall within our interval of [0,2pi) are {0, pi/3, 2pi/3, pi, 4pi/3, an 5pi/3}. whew!

Answer 37

If you view a graph, you will notice that the graph crosses at every one of these locations.

Answer 38

xp Goodness. So what did you put into your calculator? I wont ask you anything else, I know this is a lot

Answer 39

\[7\tan ^{3}x-21\tan x\]

Answer 40

Again, thank you so much for your help =) I think I'll post one more question (this one has a -7 where our 0 was, oh gawd) and call it a night

Answer 41

You're very welcome. I like it when people are curious and really seek to understand more than just their one answer.

Answer 42

^_^ that would explain you sticking around so long even after I already gave you a medal