Question

How to find x in cos2x = 3/4? Please explain in detail.

Answer 1

cos2x=2cos^2 x-1=3/4
2cos^2 x=3/4+1=7/4
cos^2 x=7/8
taking square root
cosx=0.93541
x=20.705

Answer 2

Yes, that's the value I get too. But, the question gives the interval 0 <= x <= pi. How does this work?

Answer 3

Is it \( cos^2(x) \) or \(Cos(2x)\) ?

Answer 4

It is cos(2x).

Answer 5

\[2\cos^2(x)-1=\frac{3}{4}\] may help

Answer 6

well not really
i think you can say \(2x=\cos^{-1}(\frac{3}{4})\) whatever that is, then divide by 2

Answer 7

Sorry - I should have included. In the memo, the answers are given: "x = 0.36 and x = 2.78."

Answer 8

???

Answer 9

Not sure how that can be the answer

Answer 10

I got x=20.70481105

Answer 11

I'll give the full question, but I'm not sure how it'll help.

Answer 12

This is a diff question

Answer 13

a=3 and b=-4
So the equation would be
f(x)=3-4Cos(2x)

Answer 14

No, go through the steps. I found that a = 3 and b = -4 at (i). Then, plugging those in into f(x), simplifies to cos2x = 3/4. This is where I'm stuck.

Answer 15

ohhhhhhhh i see

Answer 16

Do you have any idea where to go from here? Thanks for helping so far, by the way!

Answer 17

Okkk I got both answers.
Umm What calculator are you using?

Answer 18

I haven't got it with me here, but it's a fully adequate scientific calculator. Why? :-)

Answer 19

It's something like this: http://upload.wikimedia.org/wikipedia/commons/8/81/Casio_fx-991MS_scientific_calculator_1aa.jpg

Answer 20

ok Well I was getting the wrong answer since my calculator was on degrees setting when it should basically when you solve this on your calc make sure its on radians

Answer 21

Really? So how did you get those answers?

Answer 22

\(2\cos^2(x)-1=\large\frac{3}{4}\)

\(2\cos^2(x)=1.75\)

\(\cos^2(x)=.875\)

\(\cos(x)=\sqrt{.875} \)

\(\cos(x)= \pm .9354143467\)

\(x=\cos^{-1}( .9354143467) \text{ and } x=\cos^{-1}( .9354143467)\)

\(x=.3613671239, and, x=2.78022553\)

Answer 23

Alright! I understand everything from the first line downwards, but you have me confused right at the top.

How does \[\cos2x = \frac{ 3 }{ 4 }\] become
\[2\cos^2(x) - 1 = \frac{ 3 }{ 4 }?\]

I know the rule \[\cos^2x + \sin^2x = 1\] but I'm not sure how it applies here. You're a gem for helping me, thank you so much!

Answer 24

Its a trig identity
\(\cos(2x) = \cos^2(x) – \sin^2(x) = 1 – 2\sin^2(x) = 2\cos^2(x) – 1\)

Answer 25

I didn't know that one! I'll remember it now. Thank you SO much for your help. It feels silly just giving you a 'best response' but it's all I can do. I hope you don't mind if I ask though, but why do you do this? It's so kind and selfless, it's really great of you. Thanks again!

Answer 26

lol I used to ask loads of questions on here too so I know what you mean.
I like answering questions just to refresh my memory and its kinda fun answering questions that you know the answer to.

Answer 27

Well, you've saved my life.

Answer 28

By the way, if I close this question, will I be able to come back to it?

Answer 29

Yes on your top right hand side you will see your username in blue
Click on that and a new page will open
Then on your left hand side you will see Questions Asked in blue. Click on that and you will be able to see all the questions you have asked

Answer 30

Thanks! Another thing, sorry to bother, but that trig identity, does it apply to sin and tan as well?

Answer 31

sin(2x) = 2sin(x)cos(x)

\(tan(2x) = \large{2 tan(x) \over 1 - tan^2(x)}\)

Answer 32

http://www.purplemath.com/modules/idents.htm
I am getting all my info from here. I dont know all the trig identities off by heart.

Answer 33

I think that's all I have to ask for now. Goodnight :-)

Answer 34

Gnite :)