Question

Use a graphing calculator to solve the equation -3 cos t = 1 in the interval from

Answer 1

\[0\le \theta \le 2x\]

Answer 2

Round to the nearest hundredth and show your work.

Answer 3

is x suppose to be pi?

Answer 4

oh yeah sorry it is

Answer 5

\[\cos(t)=\frac{-1}{3} \text{ on } 0 \le t \le 2\pi \\ \text{ one solution can be found just by taking } \arccos( ) \text{ of both sides } \\ \text{ other one we can find \it by using that } \\ \text{ cosine is even and has period } 2 \pi \\ \cos(-t)=\cos(t) \\ \text{ so we have } \\ \cos(-t+2\pi)=\frac{-1}{3} \\ \text{ take } \arccos( ) \text{ of both sides and solve for } t \]

Answer 6

@freckles can you keep going cuz i really dont know how to do this

Answer 7

I fanned you and gave you a medal

Answer 8

well were you not able to the first solution?
it is just taking arccos( ) of both sides

Answer 9

\[\cos(t)=\frac{-1}{3} \text{ we could say one solution is } t=\arccos(\frac{-1}{3})\]

Answer 10

No im really confused, that's why i like when people go step by step

Answer 11

Oh I thought I gave you step by step

Answer 12

all I did there was take arccos( ) of both sides for that one solution

Answer 13

just as I said to do in the steps above

Answer 14

you did, but is that the answer?

Answer 15

that is one solution
i also said how to find the other

Answer 16

use the fact that cos is even and has period 2pi

Answer 17

so the answer is -1/3?

Answer 18

how did you get that?

Answer 19

No i'm asking you, from your step by step that's what is above that everything is equalling -1/3

Answer 20

the equation
is cos(t)=-1/3
we aren't solving for cos(t)
if we were then we would done and the answer is -1/3
we are solving for t

Answer 21

one solution is given by taking arccos( ) of both sides of the equation cos(t)=-1/3

Answer 22

ok how do we do that?

Answer 23

one solution is given by t=arccos(-1/3)

Answer 24

and then the other solution is given by using the fact that cos is even and has period 2pi

Answer 25

t=4.37255207±2πn

Answer 26

let me give you an easy example to follow:
\[\cos(\theta)=\frac{-1}{2} , \text{ where } 0 \le \theta \le 2 \pi \\ \text{ we know we are suppose to get solutions } \theta=\frac{2 \pi }{3} ,\frac{4\pi}{3} \\ \text{ we see we can get these solutions by doing : } \\ \text{ Answer one given by just takinng } \arccos( ) \text{ of both sides } \\ \theta=\arccos(\frac{-1}{2}) \\ \\ \text{ now the other solution comes from using the fact that } \\ \text{ cosine is even function and has period } 2\pi \\ \cos(-\theta+2\pi)=\frac{-1}{2} \\ \text{ solve for } \theta \text{ by first taking } \arccos( ) \text{ of both sides } \\ -\theta+2\pi=\arccos(\frac{-1}{2}) \\ \text{ now subtract } 2\pi \text{ on both sides } \\ -\theta=\arccos(\frac{-1}{2})-2\pi \\ \text{ last step multiply -1 on both sides } \\ \theta=-\arccos(\frac{-1}{2})+2\pi \\ \\ \text{ so the solutions to } \cos(\theta)=\frac{-1}{2} \text{ on } 0 \le \theta \le 2\pi \text{ are } \\ \theta=\arccos(\frac{-1}{2}) \text{ or } \theta=-\arccos(\frac{-1}{2})+2\pi \\ \text{ we can simplify these solutions } \\ \text{ We know } \arccos(\frac{-1}{2})=\frac{2\pi}{3} \\ \\\ \text{ so we have } \theta=\frac{2\pi}{3} \text{ or } \theta=-\frac{2\pi}{3}+2\pi=\frac{4\pi}{3} \\ \text{ just as we got way above just from using the unit circle }\]