Question

how do i solve for x when i have cos^3x=2?

Answer 1

does that say
\[ \cos^3(x) = 2 \]

Answer 2

yes

Answer 3

you could re-write the equation by taking the cube root of both sides
to get
\[ \cos(x) = \sqrt[3]{2} \]
what is the cube root of 2 ?

Answer 4

do you have a calculator ?

Answer 5

1.2599

Answer 6

ok, that looks good. Do you know what the biggest the cos(x) can every be ?

Answer 7

um i think the domain of cosx is -infinity to infinity

Answer 8

yes, but what is its range ?

Answer 9

but the graph of cosx is bounded by y= + and -1

Answer 10

It would be faster if you put it in a calculator to find your answer. ^_^

Answer 11

cosx is bounded by y= + and -1

yes, cos(x) ≤ 1
which means it can never by 1.2599

Answer 12

so the equation is false?

Answer 13

there are no real solutions to this equation.

Answer 14

i'm confused with the x in cosx and "bounded" referring to the y values

Answer 15

if you plot cos(x) you start with

y = cos(x)

then for every x, you figure out what y is, and plot the point (x,y)
you will get a cosine curve.
it looks like this

Answer 16

where did you get your equation \[ \cos^3(x) = 2 \]?

Answer 17

i was doing a second derivative and ended up with that

Answer 18

what was the original problem? Maybe you took a wrong turn to get that result.

Answer 19

f=sinx=tanx
i double checked my work and it seems right

Answer 20

cos^3 x = 2 was just part of the f"

Answer 21

i mean f=sinx-tanx

Answer 22

solve for x?

Answer 23

yes, OOOPS

Answer 24

but there was no real answer so yeah

Answer 25

thank phi

Answer 26

why do you have to take second derivative to solve for x?

Answer 27

and it's a function, how to "solve for x"

Answer 28

oh, i didn't mean that. i'm testing for concavity so that's why i took the second derivative of f and then solved for x

Answer 29

so that is
f(x) = sin(x) - tan(x)

and you want to solve for when f(x) = 0 ?
or you want to plot f(x) ?

Answer 30

my main goal is to plot f(x)

Answer 31

it has zeros when
sin(x) - sin(x)/cos(x) = 0 --> sinx cosx - sinx = 0 --> sinx(cosx - 1) = 0
--> sinx= 0 or cosx = 1
it goes to infinity (sign depends on sinx) when cos(x)=0

Answer 32

the first derivative f'(x) = cos(x) - 1/cos^2(x) = 0
or
cos^3(x) - 1 = 0 --> cos^3(x) = 1
which probably means you have an inflection point at x=0

Answer 33

the 2nd derivative is zero when
-sin(x) - 2 sin(x) cos^-3(x) =0
sin(x) ( cos^3(x) +2) = 0
that is zero when sin(x)=0 at x=0 so it looks like x=0 is an inflection point.