Question

solve cos^4x-1=0 for all real values of x

Answer 1

first add 1 to both sides

Answer 2

then you hav
(cos(x))^4 = 1

Answer 3

take the 4th root of each side

Answer 4

don't i have to factor the cos^4x or something O_o

Answer 5

cos(x) = +-1

Answer 6

after that do i have to find the degrees of it? that pertains to the unit circle?

Answer 7

i dont think you would need to.
once you do the root of each side
take the arccos(x)
x=arcos(1) and arcos(-1)

Answer 8

so thats the real value of x O_o\

Answer 9

yeh it should be 0˚ and 180˚

Answer 10

thats how i would do it

Answer 11

thank you so much :D!!!!

Answer 12

no prob

Answer 13

so just 0 and 180 right for the real values?

Answer 14

yeh because cos(x) has to equal 1 because there is nothing you can raise to
the 4th power to =1 except 1 and -1. and 0 and 180 are the only things that give you 1 or -1

Answer 15

I can't thank you enough WHY CANT YOU BE MORE TEACHER LOL

Answer 16

remember though, thats 0 and 180 degrees. if it asks for radians make sure to convert them

Answer 17

and your welcome

Answer 18

i have one more

Answer 19

go ahead

Answer 20

uhmmm...... 4cos^2x-3=0 for 0<x<2pi

Answer 21

ok for this one first add 3 to both sides
4(cosx)^4=3

Answer 22

then divide both sides by 4
cos(x)^4=3/4

Answer 23

then take the 4th root of 3/4

Answer 24

shouldnt it be a 2 instead of a 4?

Answer 25

im sorry yes

Answer 26

OHHHH! i gotchu, so you square root the uhmmm... 3/4 right?

Answer 27

yes, and that will give you a positive and negative value

Answer 28

so it would be cos^2x= 3/4 right? then square root that :O?

Answer 29

yes, and that will give you a positive and negative value.
then just take the arccos of those values and that should be it

Answer 30

soo it would be 3/8?

Answer 31

no actually what you end up with is when you take a square root its:
sqrt(3/4) = (3/4)^(1/2) = (3^.5)/(4^.5) = (3^.5)/2 or sqrt(3)/2

Answer 32

i have to square root 1/2 to get that O_o

Answer 33

okay thank you :)

Answer 34

no you should have to sqrt(1/2)

Answer 35

shouldn't*

Answer 36

oh okay well thank you!

Answer 37

how bout uhmm.....2x^2+3x-2=0
0 degrees <x < 360degrees

Answer 38

sqrt(3/4) = sqrt(3)/2

so youll have
cos(x) = +-sqrt(3)/2
then just take the arccos

Answer 39

ok for 2x^2+3x-2=0
0 degrees <x < 360degrees

do you know how to factor

Answer 40

i forgot to

Answer 41

one of them has -3 and 1 in it right

Answer 42

well its a bit different since you have a 2 in front of the x^2

Answer 43

where do i put the 2 then

Answer 44

Well normally you would have it broken into
(x+somehting)(x+something) where something can be positive or negative.
since you have a 2 in front itll be in the form
(2x+something)(x+something)

Answer 45

its a little tricky to actually figure out what those somethings are

Answer 46

but from your equation 2x^2 +3x -2
you know that something*somehting has to give you -2 right

Answer 47

the only things that do that are -1 and 2 or -2 and one
so you can just try those in the equation and see what works
it ends up being (2x-1)(x+2)

Answer 48

then set
2x-1= 0
and
x+2 = 0
and solve for x

Answer 49

OHHH

Answer 50

x= 1/2 and -2

Answer 51

yep

Answer 52

how aboutttt