Question

number of solutions for: sin^2+cosx+1=0 in interval (0,4pi)

Answer 1

@iPwnBunnies

Answer 2

i have a problem 'cause even when i know how to simplify the whole thing i don't get how do i get to the solutions such as x= pi/2 +kpi or something like that :|

Answer 3

@ganeshie8

Answer 4

There is an identity for sin^2 (which should be written as (sin x)^2, by the way) that would enable you to re-write the given equation in terms of the cosine function alone. What is that identity? please apply it here.

Answer 5

i need to tell you that i'm completely clueless when it comes to trigonometry >_<

Answer 6

should i add 2sinxcosx so that i can make it as (sinx +cosx)^2 or? o.O

Answer 7

I'm sorry you feel "clueless" when it comes to Trig, Salazar, but I most certainly expect that you will do something to help yourself in solving this problem. I hinted that there is an identity for (sin x)^2 and ask that you find and present that identity here.

Answer 8

Have you access to a table or list of trig identities? If not, I'd strong ly sugges you begin making up one now for continual reference.

Answer 9

i do but all i can think of using here is 1-cos^2x but then i get 2=0 which makes no sense..

Answer 10

sin^2x+cosx+1=0
(1-cos^2x)+cosx+1=0
cosx-cos^2x=0
cosx(1-cosx)=0
cosx = 0 or cosx = 1
x = pi/2 or x=0
now see in interval (0,4pi)

Answer 11

ohh right i have cosx not cos^2x so they don't shorten

Answer 12

The trig identity relating (sin x)^2 and (cos x)^2 is ... what? You fill in the blanks, please. It's important that you represent the square of the cosine as either (cos x)^2 or \[\cos^2 x\]

Answer 13

Salazar: Stop a moment. Re-read what DLS and I have typed to you. What further info need you to solve this problem yourself? Be specific, please.

Answer 14

1?
and all i need is a number of solution that work for the (0,4pi ) interval

Answer 15

solutions*

Answer 16

I see you may be experiencing an overload of help here.
You're starting with the equation (sin x)^2+cosx+1=0 and must use the identity I've been discussing to change that (sin x)^ into an equivalent function in cos x alone. What is that identity? It's essential that you know it and be able to apply it.

Using this identity, re-write (sin x)^2+cosx+1=0 in terms of cos x alone.

Answer 17

well isn't that 1-cos^2x? o.O

Answer 18

Please type that as 1 - (cos x)^2 for clarity. Yes. Use this identity to write the original equation in terms of cos x only, please.

Answer 19

1-(cos x)^2 +cos x +1=0
and i did understand what DLS wrote

Answer 20

All right. simplify 1-(cos x)^2 +cos x +1=0 now, please.

Answer 21

cos x -(cos x)^2 +2 =0

Answer 22

Good. that can be simplif. further to read (cos x)^2 - cos x - 2 = 0. Verify that this is correct or explain where you think it's wrong.

Answer 23

it's correct

Answer 24

Are you able to factor that? Hint: you might want to factor x^2 - x - 2 = 0 first, for practice.

Answer 25

(x-2)(x+1)

Answer 26

(x-2)(x+1)=. Therefore, x = ? and x=?

Answer 27

x=2 or x=-1

Answer 28

Good. Now replace that x with "cos x".

cos x = ? and cos x = ?

Answer 29

cos x =2 and cos x=-1

Answer 30

Are there any angles between 0 and 4 pi at which cos x = -1? If so, at which x? (x represents an angle).

Answer 31

If you're not sure, you could always draw or look up the graph of y = cos x and then determine where cos x = -1.

Answer 32

https://www.google.com/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=cosine%20graph

Answer 33

-pi .. -180

Answer 34

it should be -pi +2npi but i don't get how do i know it's 2npi?

Answer 35

I'd prefer you state, "cos x = -1 in only one place on the interval [0, 2pi] (or [0,360]"

cos x=-1 .... where? ... on the interval [0, 4pi]

Answer 36

720? o.O

Answer 37

Is that correct? is cos 720 = -1?

Answer 38

Again, I'd suggest you obtain the answers needed by graphing cos x on [0,4pi].

Answer 39

is cos 720 = -1? no.

Answer 40

i guess not :D

Answer 41

so, salazar, cos x = -1 at x = pi radians and then again at x = ???

Answer 42

This is an essential skill to have, so I'm going to stick to this until you come up with a solution to cos x = -1 in the interval [2pi, 4pi].

Answer 43

let me draw that

Answer 44

Just draw the graph of the cosine function on [0,2pi] and then extend it to 4pi. This will give you TWO full cycles of the cosine function. Again, this is essential to know.

Answer 45

well the line goes through 2pi

Answer 46

Salazar, as we've already seen, the solution (only solution) of cos x = - 1 on [0,pi] is pi. The cosine function is periodic with period 2pi, and thus, if we add 2pi to pi, we get 3 pi.
Thus, the roots of cos x = -1 on [0,4pi] are ... { , } (fill in the blanks)

Answer 47

(0,5pi)

Answer 48

No, afraid not; the roots are pi and 3pi: {pi, 3pi}.

Answer 49

Likewise, if you extend the interval again , to [0,6pi], the roots as {pi, 3pi, 5pi}.
How far apart are those roots?

Answer 50

what do you mean "how far apart"? how far in pi or..?

Answer 51

3p-1pi=? 5pi-3pi=?

Answer 52

oh 2 and 2

Answer 53

I appreciate your asking good questions here.

Actually, not 2 and 2, but 2pi and 2pi, OK?

Answer 54

okay

Answer 55

We've found several roots, right? the 1st positive root is x = pi.

Starting with x = pi, add 2pi, then add 4pi , then add 6 pi.
what do you get?

pi, 3pi, ..., ...,

Answer 56

Sal? pi, 3pi, ? , ? , ....

Answer 57

Sorry to make you repeat this over and over, but I want you to see the pattern that is developing.

Answer 58

okay pi,3pi,7pi,13pi

Answer 59

Wouldn't that be pi, 3pi, 5pi, 7pi, 9pi, ...?

Answer 60

Why 7pi, 13pi, with no 9pi, no 11pi?

Answer 61

you said ad 4 and then 6...

Answer 62

add*

Answer 63

But anyway. So, if you choose x = -1 as your beginning root, the next root is x=-1+2pi; the next root is -1+4pi, the next is -1+6pi, and the next is .... ??

Answer 64

-1 + 8pi?

Answer 65

Exactly. And therefore, your general solution is x = -1 (plus or minus)n*2pi, where n=0, plus or minus 1, plus or minus 2, and so on.

If you want the solutions ONLY on [0,4pi], let n=0 and 1 and you will end up with
pi+0*2pi = pi
pi+1*2pi = pi + 2pi = 3pi (which we already knew).

Answer 66

Does this conversation answer your questions / concerns?

Answer 67

okay okay , got it xD
thank you :)

Answer 68

My pleasure. Thanks for sticking with me and not complaining ab out my going on and on. I truly wanted you to undrstand the concepts behind this problem.

Answer 69

np :D