Question

Solve for x. EQUATION BELOW!

Answer 1

\[\frac{ 7 }{ 9 }pi\sin(\frac{ 7 }{ 36 }\pi x)\cos(\frac{ 7 }{ 36 }\pi x)=0\]

Answer 2

i have it down to :

Answer 3

\[\sin(7/36 \pi x) *\cos(7/36 \pi x) =0\]

Answer 4

I am stuck totally at this point

Answer 5

Use the following trigonometric identity:

sin 2x = 2 (sin x) (cos x)

Answer 6

I just got that by dividing 0 by 7/36pi x . I hope that was legal

Answer 7

But there isn't a two out front...

Answer 8

even if i back up and don't divide over the front constant...

Answer 9

You have to "manufacture" a 2 by utilizing (2)(1/2) = 1

Answer 10

so would i do \[\sin (\frac{ 7 }{ 9 }\pi * \frac{ 7 }{ 36 }\pi x) = 0\]

Answer 11

oh, ok. wait what do you mean by manufacture? Sorry i'm being dumb tonight

Answer 12

i have figured that \[\frac{ 7 }{ 36 }\pi * 3.274044544 = 2\]

Answer 13

does that do me any good?

Answer 14

You're not dumb at all. You're actually one of the better students here because you're trying to do your own work, which is commendable, but is really the way it is supposed to be anyway, because no one learns without trying. And you're doing an admirable job of trying. You're getting close, also.

Your "x" here is 7(pi)x/36 so you know how to double that. And as for there not being a "2" in front, I'll rewrite my identity:

(1/2) (sin 2x) = (sin x) (cos x)

There, now the 2 on one side becomes the 1/2 on the other side.

Answer 15

so my equation is now:

Answer 16

And you can keep your calculations in terms of pi. No real need to use that long decimal number, really.

Answer 17

\[(\frac{ 7 }{ 18 })(\sin \frac{ 7 }{ 18 }\pi x)=0\]

Answer 18

that is backing up a step before I didivded that leading term out

Answer 19

but i guess that doesn't make a difference and I end up with \[\sin \frac{ 7 }{ 18 }\pi x = 0\]

Answer 20

You're doing really well now, but did you leave out a "pi"? It looked like you had one outside of the trig functions in your original problem statement.

Answer 21

oh your right, there was a a pi with that leading 7/18, but that doesn't matter right because im just gonna bring it over and it will go away. right?

Answer 22

Yes, that is absolutely correct. It wouldn't matter, because you can divide 0 by that pi just like you did with numeric. So now you have only one trig function to deal with and we know that

sin a = 0

at a = 0 or pi or 2pi or 3 pi, etc.

So, you just have to solve for "a".

Answer 23

and how on earth would i do that? I just got a little lost on your last step I think...Sorry

Answer 24

how would I solve for a?

Answer 25

Look at your unit circle. Where does \(\sin a = 0\)?

Answer 26

np. What I did was put into your mind just where you have

"sine of an angle is 0"

So, you have :

sin (7)(pi)(x)/18 = 0

So, you have to see all places where:

(7)(x)/18 <----- without the "pi"

equals 0, 1, 2, 3, 4, 5, etc

and that is at:

x = 0, 18/7, 36/7, 54/7, etc.

Is that making sense now?

Answer 27

i mean i get that, but i am doing sin(7/18 pi x) =0 , not sinx=0

Answer 28

do I just multiply my answers, aka 0 and pi by 7/18 pi?

Answer 29

what do you mean by " See all the places where 7x/18 = 0,1,2, etc" ?

Answer 30

am i plugging 7x/18 into sin?

Answer 31

\[\sin 0 = 0\]\[\sin\pi = 0\]\[\sin 2\pi = 0\]\[\sin n\pi =0\]if \(n\) is an integer

Answer 32

Now, a lot of times, a problem will ask for a solution "in the first quadrant" or "in the unit circle". You don't have that in your problem statement, so we can add that logically, to make a sensible answer, because it all comes back to :

sin 0 = 0
sin pi = 0
sin 2pi = 0
sin 3pi = 0 etc.

Answer 33

I think mine is for the unit circle. What i'm trying to do is find the critical points of a function and what we are doign is solving the derivative for 0 to find those critical points. So i'm assuming from 0,2pi is my interval

Answer 34

ok, so I think i get what you are saying, but i'm not sure how i"m suposed to be using it.

Answer 35

That is a very logical assumption, but one shouldn't make too many or unnecessary assumptions. If that is your interval, then take your first 3 values of "x".

Answer 36

I know that sin x = 0 when x = 0, pi, 2pi

Answer 37

first three values of x? what do you mean?

Answer 38

so do I do 0*7x/18pi?

Answer 39

x = 0, 18/7, and 36/7 because that will give you your sin 0, sin pi, and sin 2pi.

Answer 40

Oh, ok.. I think i am following. I guess

Answer 41

wait, no, how did you get those x values?

Answer 42

Any value of x where \[\frac{7 x}{18} = n\]where \(n\) is an integer, \(\sin (7\pi x/18) = 0\).

Answer 43

Don't leave with a fuzzy warm feeling. Especially the "fuzzy". Ask anything. But soncentrate on what you know:

sin 0 = 0
sin pi = 0
sin 2 pi = 0

Answer 44

If you rearrange it, \[x = 18n/7\]\(n\) is an integer

Answer 45

so i do 0* 7x/18 pi?

Answer 46

and then pi * 7x/19 pi?

Answer 47

You know those things. Now, all you are doing is substituting expressions in "x" that will come down to multiples of pi for which you know values.

Answer 48

Ok, didn't get any of that last post tcarrloll

Answer 49

so i'm trying to get my 7x/18 pi to equal 0, pi, and 2pi?

Answer 50

That "(7)(pi)/9" in the original problem is extraneous for finding where you want the rest to equal 0.

Yes, to that last question.

Answer 51

so i set \[\frac{ 7x }{ 18}\pi=0 \]

Answer 52

Let's make a little table:

n 18n/7 7 pi x/18
-2 -36/7 7*pi*(-36/7)*1/18 = -2 pi
-1 -18/7 7*pi(-18/7)*1/18 = -1 pi
0 0 7*pi(0)*1/18 = 0
1 18/7 7*pi(18/7)*1/18) = pi
2 36/7 7*pi(36/7)*1/18) = 2 pi
etc.

Answer 53

and solve for x? and do the same setting that equal to pi and then 2 pi?

Answer 54

I have no idea what you are doing with your table whpalmer. what are each of the columns?

Answer 55

So, you are getting 7x/18 to equal 0, 1, 2. The pi is already there as a multiplier. I think you are over-thinking this now.

Answer 56

I really think i may be. I am so lost feeling

Answer 57

Lets back up a bit I guess

Answer 58

I need to find values for x, so that 7x/18 = 0, pi, and 2pi correct?

Answer 59

why am i ignoring that pi with the 7x/18?

Answer 60

no, so that 7x/18 = 0, 1, 2, etc.

7x * pi/ 18 = 0pi, 1pi, 2pi

Answer 61

Good. More simple: (simpler?)

sin (a)(pi) = 0

a = 0, 1, 2

Look at that for a moment.

Answer 62

oh ok, i think it just clicked!

Answer 63

It might be helpful to think about "manufacturing" that 2 way back when - you're really doing the same thing here...

Answer 64

we wanna transform the equation into 3 seperate equations that we know the answer for. correct?

Answer 65

All I did there was work with a simpler multiplier of "pi". That's all you are doing with that (7)(x)/18

Answer 66

so I do \[\frac{ 7 }{ 18 }\pi x = \pi \]

Answer 67

and solve for x,

Answer 68

You can transform into 3 equations, but that is splitting things up when you don't have to. That unnecessarily complicates things. It's like wpalmer said. Look at the unit circle, going counter-clockwise from 0 to 2pi and then stop.

Yes, just solve for "x" to get 0, 1, 2

Answer 69

so i get x=18/7

Answer 70

then x=36/18?

Answer 71

Because the "pi" is already there. Yes, you get 18/7 to get to "pi" And you also get 36/7 to get to 2pi. And you get 0 to get to 0 x pi.

Answer 72

and 0
so my three solutions are x = 0, 18/7, and 36/18?

Answer 73

i meant over 7 for that last one

Answer 74

No. Your "x" is multiples of 18/7.

Answer 75

oh god lol

Answer 76

Yes, you have it now. Good job! I suggest going over this once when you have time.

Answer 77

I am just told to find the critical points of a function and I know to do that I must solve the derivative for 0 and those values of x are my critical points. I am then to graph the critical points. I think we're over complicating it now.

Answer 78

Who is "we" here? lol :-)

Answer 79

so x=0,18/7,36/7 are my critical points, correct?

Answer 80

Yes

Answer 81

Lol just me. Spreading the blame to make me feel better ha You have been a great great help! above and beyond.

Answer 82

Thanks so much!. I get so tripped up with trig functions. I did the other problems just fine

Answer 83

I just gave wpalmer a medal too, because he was a big help too. Way to go Wpalmer!

Answer 84

I'd fan you if I could, but it seems to be messing up a bit.

Answer 85

Yes, something is wrong with Openstudy right now. I have been having the worst time typing. Letters keep getting chopped off.

Answer 86

well even greater thanks. Have a great night!!

Answer 87

You, too! Always! thx for medal!