Question

http://prntscr.com/9av46p
Do I not divide the 2 in the cos function out to get the answer?? o-0

Answer 1

how did you get those two answers?

Answer 2

hopefully you agree that if
\[\Large \cos(2x) = \frac{\sqrt{3}}{2}\]
then
\[\Large 2x = \frac{\pi}{6} + 2\pi*n \ \ \text{ or } \ \ 2x = -\frac{\pi}{6} + 2\pi*n\]

Answer 3

I don't exactly remember but I think it involved dividing the 2 out of the cos2(x)

Answer 4

do you see how I got \[\Large 2x = \frac{\pi}{6} + 2\pi*n \ \ \text{ or } \ \ 2x = -\frac{\pi}{6} + 2\pi*n\]

Answer 5

Um... no, sorry

Answer 6

use this unit circle

https://upload.wikimedia.org/wikipedia/commons/thumb/4/4c/Unit_circle_angles_color.svg/2000px-Unit_circle_angles_color.svg.png

and tell me what angle corresponds to the point with x coordinate \(\Large \frac{\sqrt{3}}{2}\)

Answer 7

there are 2 such angles

Answer 8

Okay. I may take awhile, so you don't have to wait (sorry, multitasking) !

Answer 9

that's fine

Answer 10

cosine is a function of an angle

that angle measure is a function of x, you cant take 4 out

Answer 11

thats 4*an angle, not same as 4*cosine(angle)

Answer 12

I'm still lost...

Answer 13

Wait is it cos(x), find x then div by 2?

Answer 14

\[\cos(\theta) = \frac{ \sqrt{3} }{ 2 }\]

is given, and theta, is defined as in terms of an x

\[\theta = 2*x\]

Answer 15

so theta/2 to get x

Answer 16

A*sin(B*x) , B changes the period of the wave

Answer 17

Wait what?

Answer 18

What I understood from that: (arcos (sqrt 3 / 2) div by 2

Answer 19

\[\cos(\theta) = \frac{ \sqrt{3} }{ 2 }\]


for theta is pi/6, and -pi/6, need both between 0 and 2pi


rewrite -pi/6 as 2pi - pi/6 = 5pi/6

Answer 20

Okay...wait why the negative pi/6?

Answer 21

there are the two angles for THeta

pi/6 and 5pi/6


theta is a function of x,

theta = 2*x

so the x values they want are actual x = theta/2

Answer 22

Yes... okay, I get that

Answer 23

half of both those angles, x = pi/12 , x = 5pi/12

Answer 24

Yes

Answer 25

oh so my mistake was the 22pi/12?
I got that off Mathway... lol must've mistyped

Answer 26

?

Answer 27

maybe the negative because when you take the inverse of a function that isnt 1-to-1, it wont work. i forget the names of those values , principle angles , or restricted, idk

Answer 28

you have to restrict cosine to angles, or domain from -1 to +1

Answer 29

eh.. okay

Answer 30

think of a reflection over the line y=x, wont work for cos, have to only take a piece from angles 0 to pi, then you can get inverse y=arccos(x) or x = cos(y)

Answer 31

what...?

Answer 32

so 5pi/6 doesn't work?

Answer 33

nvm, if you havent learned about the inverse trig domain and ranges yet

Answer 34

I have actually but I lost you around "reflection" lol

Answer 35

oh, yeah remember finding inverses of algebra functions f^-1

you were told to switch the y and the x around, then solve for y, that is the inverse function

Answer 36

Yeah

Answer 37

same with the trig things, but you have to be careful...

y = cos(x)

inverse cosine--
x = cos(y), and is same as , y = cos^-1(x)

Answer 38

o-o where does inverse apply here?

Answer 39

but you have to check the domains more careful,

trig functions are not 1 to 1 functions, many theta values for any value for cos(theta)

because they are periodic and repeat

Answer 40

the thing has to be 1x per 1y value all the time, 1-to-1 function , to do the inverse

Answer 41

??

Answer 42

Ok.

Answer 43

for any x value you pick, only one y can happen, unlike many angles giving same y value for trig things

Answer 44

Actually I need to do other problems, let's just come back to this later

Answer 45

so you limit the domain of cos(theta) to angles 0 to pi, in order to reason the inverse cosine


cosine by itself has any number for th edomain

Answer 46

ok, try to reflect cos(x) over the line y = x

each point on cos(x) jumps over y=x perpendicularly like a mirror

you can only do that if you take cos(x) from just 0 to pi