Question

I don't know how to get it to where there is just one cos for the whole equation. I thought maybe I could put one on each side & sqrt it but I don't think that's going to work.. Any ideas?

(picture below)

Answer 1

start with
\[\cos(2x)=2\cos^2(x)-1\]

Answer 2

How did you get the 2 in front of the cos^2x?

Answer 3

making the left hand side
\[3\cos^2(x)-1\]

Answer 4

it is one version of the "double angle" formula for cosine
it says
\[\cos(2x)=2\cos^2(x)-1\]

Answer 5

So that's now the equation your working with? Okay, sounds good. But I'm not sure how you go to the 3cos^2x-1 part.. & thanks for helping again, I really appreciate it.

Answer 6

ok the left hand side is \(\cos(2x)+\cos^2(x)\) i replaced \(\cos(2x)\) by \(2\cos^2(x)-1\) to get
\[3\cos^2(x)-1=1\] but now that i look at it, that might not be the way to go
maybe a better idea is to go the other way around, preplace \(\cos^2(x)\) by
\[\frac{\cos(2x)+1}{2}\]

Answer 7

lets try it that way. but you have to use one form of the double angle identity, there is no avoiding it, because one input is \(x\) and the other is \(2x\)

Answer 8

Oh, I see what you did now. & okay, so that would make it 2cosx + (cos2x+1)/2

Answer 9

yes

Answer 10

that gives
\[\frac{3}{2}\cos(2x)+\frac{1}{2}=1\] now we can solve i think

Answer 11

I got as far as cos2x=1/3 but I'm not sure how to get rid of the 2 in cos2x

Answer 12

\[\frac{3}{2}\cos(2x)=\frac{1}{2}\]
\[\cos(2x)=\frac{1}{3}\] oh damn i am stuck

Answer 13

is cos2x the same as cos^2x?

Answer 14

Wait, nevermind. Of course not

Answer 15

oh no, they are different

Answer 16

Hmmm. Well I'm stuck..

Answer 17

me too but i am sure we can do it
one thing to do is check which one works

Answer 18

Okay, I'll try that. Hold on:)

Answer 19

None of them worked for me..

Answer 20

i think none of these are exact
there is something odd going on here

Answer 21

if we take \(\cos^{-1}(\frac{1}{3})\)you get \(70.53\) so if \(2x)=70.53\) then \(x=35\) but not really, only rounded

Answer 22

Hmm. I'm not sure, I might just have to ask my teacher.. But I really do appreciate all the help you've given me:)

Answer 23

gotta run, good luck
i think there is something strange here, none of these answer are really correct

Answer 24

yw