Question

Solve for x EQUATION BELOW

Answer 1

\[\frac{ -49 }{ 864 }\pi^2(\sin \frac{ 7 }{ 36 }\pi x) -3\sin(\frac{ 7 }{ 12 }\pi x) =0\]

Answer 2

dividing over gives just :

\[\sin(\frac{ 7 }{ 36 } \pi x) -3\sin(\frac{ 7 }{ 12 }\pi x)=0\]

Answer 3

It has the form sinx-3sin3x=0

Answer 4

does it? lol

Answer 5

7/12 = 3* 7/36

Answer 6

oh indeed it does. What does that do for us though? lol

Answer 7

Don't know if it does :/

Answer 8

exactly lol

Answer 9

Could try to use sum formula for sin: sin3x=sin(2x+x)=sin2xcosx+cos2xsinx

Answer 10

i tried seeing what wolfram explained it as and it used some crazy stuff that I totally didn't understand like " weirstrass substitution"

Answer 11

maybe we could use that

Answer 12

OK, I think I am on to something:

Answer 13

hooray!

Answer 14

ust leave out the 7/36 pi stuff for now:

Then:
sinx=3sin3x
sinx=3(sin2xcosx+cos2xsinx). Divide everything by sinx:
1=3(sin2xcosx/sinx+cos2x). Use sin2x=2sinxcosx formula:

Answer 15

1=3(sin2xcosx/sinx+cos2x)

1/3=2sinxcosx/sinx +2cos²x-1
1/3=2cosx+2cos²x-1
2cos²x+2cosx-4/3=0

This is a quadratic equation with cosx as variable.

Answer 16

I am not really following much of this, but if you come to a good answer I"ll look back at it and try to see your logic. So don't pause for me! lol

Answer 17

I know you can!

Change it (by dividing by 2 and multiplying with 3) to:

cos²x+cosx-2=0

Using quadratic formula:

\[\cos x=\frac{ -1 \pm \sqrt{1+8} }{ 2 }=-\frac{ 1 }{ 2 }\pm \frac{ 3 }{ 2 }\]

Answer 18

oh, ok. I sorta see where you're coming from. I actually just emailed my teacher to make sure this wasn't a typo or something because solving this seems WAY out of my league

Answer 19

Now, one of these numbers is -2, and that is not a valid cos-value.
The other one is 1, and gives us: (after putting back in the 7/36 pi stuff):\[\cos(\frac{ 7 }{ 36 }\pi x)=1\]

Answer 20

That is a good idea. If it is a typo, we are doing all this just for nothi... fun. :)

Answer 21

BTW, it seems now that we have come to an equation you can solve...

Answer 22

if you say so, lol i would really appreaciate if you could maybe follow through with your thought. I have a feeling that I really am gonna have to do this

Answer 23

I'm not sure what you mean.. I'm from Holland, and mostly I speak Dutch :p

Answer 24

Oh, ok. Sorry.

Answer 25

Could you please go ahead and solve the equation out if you could.

Answer 26

Yes: \[\cos(\frac{ 7 }{ 36 }\pi x)=1 \Leftrightarrow \frac{7}{36}\pi x = 0\](of cours you could add 2kpi to get all the solutions, but idk if you want them all).

So this leads only to x=0. Sounds too good to be true?

Answer 27

I"m gonna go with it! Thank you so so so very much as usual

Answer 28

YW! Let me hear when you have talked about it with your teacher!

Answer 29

@gabie1121: I made a mistake :(

Somewhere up I had:

1=3(sin2xcosx/sinx+cos2x)
1/3=2sinxcosx/sinx +2cos²x-1 <<<< this is wrong!

It is:

1/3=2sinxcosxcosx/sinx +2cos²x-1
1/3=2cos²x+2cos²x-1

This will be:

4cos²x=4/3
cos²x=1/3, so\[\cos x = \pm \sqrt{\frac{1}{3}}=\pm \frac{1}{3}\sqrt{3}\]

This cannot be solved easily.

But there is also good news: because we divided by sinx along the way, we left out the solution sinx=0, which means x=0, pi, 2pi...

So our solutions x=0 is still there!

Here is a drawing of the grapgh of y=sinx-3sin3x:

YOu can see the k*pi solutions along with the difficult ones...

Answer 30

Was the \(\large \frac{-49}{864}\pi^2\) multiplying the first term, or the entire equation on the left? Your brackets are a little funky so it's hard to tell.

Answer 31

AFAIK, it was the entire equation, that's why it was cancelled.

Answer 32

yeah, it was the entire equation