Question

Solve for x EQUATIONS BELOW

Answer 1

\[1+\cos(\frac{ 7 }{ 36 } \pi x)=0 \]

\[1-\cos(\frac{ 7 }{ 36 } \pi x)=0\]

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

Answer 2

I know cos=0 at pi/2

Answer 3

and cosx =1 at 0 and -1 at pi

Answer 4

do I just set up like this: \[\frac{ 7 }{ 36 } \pi x = -1 \] and solve for x?

Answer 5

and then do similiar for the others?

Answer 6

OK, take the first one: make it \(\cos (\frac{7}{36}\pi x)=-1\), then, as you said, \[\frac{ 7 }{ 36 }\pi x=\pi + 2k \pi\]

Answer 7

well i'm solving technically to get critical points to graph, so I think i don't need to be super general and cover them all

Answer 8

OK, so you can stay within on period. We'll lose the k-part later.
To solve this, we can divide everything by pi:\[\frac{ 7 }{ 36 }x=1+2k\]
What to do next?

Answer 9

wouldn't my answer just be 36/7? for x

Answer 10

It would, because you do not have to care about the other solutions!

Answer 11

awesome.

Answer 12

can i have you help me with another soemthing real quick why i have you?

Answer 13

You've got me, girl :D

Answer 14

hooray! ok so the problem i am working is using a function and finding the critical and inflection points. What we were just doing was helping to find the critical points

Answer 15

The origianl function is :

Answer 16

\[2.25+2\sin^3(\frac{ 7 }{ 36 }\pi x)\]

Answer 17

i found the first derivative to be:

Answer 18

\[\frac{ 7 }{ 6 }\pi \sin^2(\frac{ 7 }{ 36 }\pi x)*\cos \frac{ 7 }{ 36 }\pi x)\]

Answer 19

That's OK!

Answer 20

thats the equation that you stepped in the last bits of solving that for zero to find the critical points. Now i have to find teh second derivative and solve that for zero to get my inflection points. ANd here is where i need some help

Answer 21

Wolfram alpha says that the second derivative is (i'll work on getting to it on my own in a bit):

Answer 22

wait.. maybe i should try and get it on my own first.

Answer 23

OK, I'll wait...

Answer 24

anyway my problem is solving it and thats what i need you for. I dont wanna make you wait

Answer 25

wolfram says its : \[\frac{ -49 }{ 864 }\pi^2(\sin(\frac{ 7 }{ 36 } \pi x)-3\sin(\frac{ 7 }{ 12 } \pi x))\]

Answer 26

which i can see making sense by the product rule and then using an identity or two to get to that point. Now i gotta solve that beast for 0

Answer 27

Well, I'm not waiting, I'm finding the second derivative myself; what a monster!

Answer 28

i know at least that i can ignore those opening multipliers cuz they'll just disappear when i divide them over

Answer 29

so i'm stuck with \[\sin(\frac{ 7 }{ 36 }\pi x) - 3\sin(\frac{ 7 }{ 12 }\pi x)=0\]

Answer 30

Perhaps i should make a new question out of this...

Answer 31

OK!