Question

cos(4x)-cos(6x)=0 find all values 0 11 years ago

Answer 1

you could use sum to product or what it is called

Answer 2

how did you achieve cos(4x)=4cosx there is not such thing

Answer 3

I dont think you can do that @ageta

Answer 4

cos(4x)-cos(6x)=2sin(5x)sinx

Answer 5

set that equal to zero
i used sum to product if you don't know i shall explain...

Answer 6

clear?

Answer 7

Yes but this leaves me in a hole almost as large.. I am trying to get it down to a single sin(x) or cos(x) in order to find the angles that satisfy the question

Answer 8

what do you mean?
you 2sin(5x)sinx=0 ===>sin5x=0 or sinx=0

Answer 9

yes?

Answer 10

it is not all the time a single sin or cos as you think

Answer 11

you could have tanxsin5xcos3x=0
this is very legitimate

Answer 12

So how do I use sin(5x) to find the angles?

Answer 13

if you have product of terms equal to zero it means that each term
can equal zero

Answer 14

like when you have (x+2)(x-3)(x-1)=0 =====>x=-2 or x=3 or x=1
the same thing applies to trig (it is just algebra again)

Answer 15

sin(5x)=0 you ask what angle in the range 0 to 2pi gives us 0 when applying sin to it
that would be 0, pi, 2pi

Answer 16

yes?

Answer 17

so the 5x is not significant?

Answer 18

5x is a whole quantity by it self it is acting exactly like x

Answer 19

it is not about significance! you have 5x you need to solve for x only
so you got as i said 5x=0 or 5x=pi or 5x=2pi then solve for x

Answer 20

that's only for sin5x=0 after you move to sinx=0

Answer 21

eh i'm sorry 0 and 2pi are not included actually i didn't pay attention
i thought it was 0<=x<=360

Answer 22

so it's only 5x=pi

Answer 23

The answers for the angles that are results of cos(4x)-cos(6x)= 0 are: 0, pi/5, 2pi/5, 3pi/5, 4pi/5, pi, 6pi/5, 7pi/5, 8pi/5, 9pi/5,

Answer 24

I meant 0<x<360
- -

Answer 25

However i still do not see how one would arrive at all of those angles

Answer 26

don't worry about options
you have 5x=pi ====>x=pi/5 that's one solution

Answer 27

hmm sum of them might not be answers
since we are only taking from 0 to 360

Answer 28

These are the answers in the textbook, I was just trying to find out how to arrive at them

Answer 29

for sin5x we found x=pi/5 for \(\large x\in(0,2\pi)\)

Answer 30

I think the way to go is cos(2x+2x) -cos(4x+2x) and just to finish that out

Answer 31

now move on to sinx=0 ===> x=pi for \(\large x\in(0,2\pi)\)

Answer 32

to me it appear there only two solution to \(\large cos4x-cos6x=0\) when
\(0<x<360 ~or~ 0<x<2\pi\)

Answer 33

unless if you did a mistake in typing the range of the solutions

Answer 34

nope thats gread thank you very much!

Answer 35

No problem!
never look at answer options they just confuse you

Answer 36

focus on solving first

Answer 37

and you need to get you sum to product formula memorized if you can't bring them back by any other technique they are very important

Answer 38

http://colalg.math.csusb.edu/~devel/IT/main/m05_identities/src/s04_sumtoproduct.html
this might help