Question

Help :)

cos(3x)=-1

Part 1) Solve this equation for 3x, finding all values on the interval [0,2π] that satisfy the equation.

Part II) Use algebra to find all values of x between 0 and 2π that satisfy this equation.

Answer 1

so how do you undo cosine?

Answer 2

umm...would i have to put 3x by itself

Answer 3

to solve for 3x

Answer 4

yes, but first how do you undo cosine? just like how do you "undo" multiplication

Answer 5

you undo multiplication with division, sooo cosine with secant?

Answer 6

hellooo? I still kind of need help...anyone...^-^

Answer 7

uhm not quite you use inverses

Answer 8

ohh yeah, so cos^-1 or arccos

Answer 9

yup to both sides

Answer 10

for when you compute, so do the first step

Answer 11

cos(3x)=-1

cos(3x)+1=0

is this correct

Answer 12

I wouldn't start it that way because you cannot undo the cosine now

Answer 13

your first step should be to get rid of the cosine

Answer 14

okay...umm..

\[\cos(3x)/\cos^-1=-1/\cos^-1\]

Answer 15

3x= -1/cos^-1

Answer 16

That doesn't seem right...I dont know what im doing! aaaaaaaaa

Answer 17

dont i have to set the equation to equal 0 or is that the case for this equation

Answer 18

uhm, not quite, so cosine is a function like logarithms. ie lnx=1 e^lnx=e^1 x=e^1 or e^x=1 ln(e^x)=ln(1) x=0

Answer 19

or what is f(f^-1(x))=?

Answer 20

you can't divide by a function, you apply it's inverse to both sides to undo it

Answer 21

im kind of going off of this equation

2cosx-sqrt2=0 at the end i got cosx= sqrt2/2 the found two solutions. i think thats what the question wants me to do

Answer 22

huh? now i'm confused

Answer 23

ok.

2cosx-sqrt 2=0
2cosx-sqrt2 + sqrt2= sqrt2
2cosx/2 = sqrt2/2
cosx=sqrt2/2, then i found pi/4 and 7pi/4 to be solutions

Answer 24

but this equation is much more complicated in the sense that i cant just do those simple steps

Answer 25

hmm

Answer 26

on (0,pi)?

Answer 27

(0,2pi)*

Answer 28

yes.

Answer 29

because I'm having a brain fart, do those equal 45 deg

Answer 30

i plugged in this function in this website, but didnt too much help
http://www.wolframalpha.com/input/?i=+cos%283x%29%3D-1+solve+

Answer 31

I was helping you on that

Answer 32

let's do a similar ex

Answer 33

let's say we have sin(2u)=1

Answer 34

my first step is to get rid of the sine

Answer 35

okay, thats where im confused

All the equations i have done with trig functions dont involve getting rid of cosine or sine and such

Answer 36

so i dont know how to do that

Answer 37

so i do this: \[sin^{-1}[sin(2u)]=sin^{-1}(1)\]
\[\not{sin^{-1}[\not{sin}}(2u)]=sin^{-1}(1)\]
\[(2u)=sin^{-1}(1)\]

Answer 38

that's supposed to be cancelling the sin and sin^-1 since they equal 1

Answer 39

isnt cos(3x)=-1 already simplfied
for ex.

If i would simplify 2cos(7x)+1=0 i would get cos(7x)=-1/2

Answer 40

not yet

Answer 41

if it's x= then it's simplified

Answer 42

you want x by itself

Answer 43

oh, i see how you did that

Answer 44

i would get 3x= cos^-1(-1)

Answer 45

so now, in mine we will solve for u

Answer 46

but yes you will eventually get that. It's important to show your steps though

Answer 47

i got pi/3

yes, i will

Answer 48

since 3x = pi

3x/3=pi/3

x=pi/3

Answer 49

cos^-1(-1)= pi

thank for your help

Answer 50

@_@

Answer 51

i can buy that

Answer 52

np

Answer 53

medal?