Question

PLEASE HELP! Trig quiz last question. Question attached in photo.

Answer 1

I have an idea for you, say\[Let "Cosθ"=a\]simplify it first and tell me what you get after simplifying.

I hope this makes it look easier, if you solve for a.

Answer 2

my bad, actually say\[Let "Cos3θ"=a\]I mean either way.

Answer 3

oh okay now i undestand ok let me see
\[a ^{2}-6a+4=0\]

Answer 4

I did this already but with x, and i got as far as \[3 plus or minus \sqrt{5}\]

Answer 5

Do this now, show me your steps.

Answer 6

I mean do it with a.

Answer 7

okay. a=1 b=-6 and c=4

a= - (-6) +/- \[\sqrt{(-6)^{2}-4(1)(4)} \over 2(1)\]

Answer 8

\[6 \pm \sqrt{36-4(4)}/\over2\]

Answer 9

\[6\pm \sqrt{20}\over2\]

Answer 10

yes, but this is not it.

Answer 11

\[6\pm2\sqrt{5}\over2\]

Answer 12

Yep, and it will be?

Answer 13

\[3\pm \sqrt{5}\]

Answer 14

Yes.

Answer 15

\[a=3+\sqrt{5}\]

Answer 16

\[Cos3θ=3+\sqrt{5}\]\[and\]\[Cos3θ=3-\sqrt{5}\]Solve for θ in each.

Answer 17

this is where i am stuck. Do i do arc cos first then divide by 3?

Answer 18

we can use the same idea again Let _ = _

Answer 19

\[Let "3θ"=x\]

Answer 20

let 3 theta = a?

Answer 21

Yep, you said it also.

Answer 22

So,\[Cosa=3-\sqrt{5}\]\[and\]\[Cosa=3+\sqrt{5}\]
you can use an approximate value for \[\sqrt{5}\]

Answer 23

well, \[cosa=3+\sqrt{5}\] does not work because it is not within the range of cosine

Answer 24

\[a=\cos^{-1} (3-\sqrt{5})=40.1879\]

Answer 25

\[3\theta=40.1879\]
\[\theta=13.396\]

Answer 26

hold on, I would re-put it, substituting an approximate value of sqrt5, before doing inverse cosine.
\[\sqrt{5}=2.24\]\[SO,\]\[Cosa=.76\]\[and\]\[Cosa=5.24\]Obviously the second one doesn't work.

Answer 27

yes because it is outside the range of cosine

Answer 28

Good, so now we have\[Cos ^{-1}(.76)=?\]divide this by 3 when you get the result.

Answer 29

13.5

Answer 30

Is that the final, after dividing by 3?

Answer 31

yes

Answer 32

(\[Good!\])

Answer 33

now I don't know what to do.

Answer 34

or is that the answer?

Answer 35

Well you will have 2 answers.

Answer 36

You will have what you got, then you will have another positive value, --> that+180.

Answer 37

why?

Answer 38

Why, 2?

Answer 39

why do we add 180. why quadrant 3?

Answer 40

cosine is positive in quadrant 1 and 4. so i thought 13.4 (the answer when you dont round \[\sqrt{5}\]) and 360-13.4 which is 346.6 but this is incorrect :/

Answer 41

No, because you are looking for all positive values btw 0 and 360.
I didn't say it is Quadrant 3, but right, b/c

you are adding the angle to 180, not 180 to the angle.