Question

please help... FAN AND MEDAL

Answer 1

@freckles

Answer 2

which one do you looking at?

Answer 3

all of them?

Answer 4

yes all

Answer 5

ok well lets look at:
\[\cos(2x)=\frac{1}{\sqrt{2}} \\ \text{ or also known as } \\ \cos(2x)=\frac{\sqrt{2}}{2}\]
Before we solve this one, do you know how to solve:
\[\cos(\theta)=\frac{\sqrt{2}}{2}\]
also are we solving in a specific interval ?

Answer 6

the interval is between 0 and 360

Answer 7

45 and 315

Answer 8

ok if we have
\[0<x<360 \\ \text{ and we \let } theta=2x \\ \text{ then } x=\frac{\theta}{2} \\ \text{ and so we have } \\ 0<\frac{\theta}{2}<360 \\ \text{ multiplying both sides by 2 we have } \\ 0< \theta<720 \\ \text{ so you said } \cos(\theta)=\frac{\sqrt{2}}{2} \text{ has solutions } \\ \theta=45^o,315^o, 45^o+360^o,315+360^o \\ \text{ after simplifying } \\ \theta=45^o,315^o,405^o,675\]

Answer 9

now recall
\[\theta=2x \\ \text{ so we have } \\ 2x=45^o ,315^o,405^o,675^o\]

Answer 10

just divide both sides by 2

Answer 11

so 22.2?

Answer 12

157.5?

Answer 13

ok and two more solutions

Answer 14

\[x=\frac{45^o}{2},\frac{315}{2}^o,\frac{405^o}{2},\frac{675^o}{2}\]

Answer 15

we had to solve cos(theta)=sqrt(2)/2 in (0,720) since x was between 0 and 360

Answer 16

theta was 2 times more than x

Answer 17

ahh gotcha.. and thats the answer?

Answer 18

yes those 4 solutions right there at mentioned are the answers

Answer 19

is that considered double identity angles?

Answer 20

we didn't use the double angle identity but we could have...
\[\cos(2x)=\cos^2(x)-(1-\cos^2(x)) \\ \cos(2x)=2\cos^2(x)-1 \\ \text{ so we have } 2\cos^2(x)-1=\frac{\sqrt{2}}{2} \\ 2 \cos^2(x)=\frac{\sqrt{2}}{2}+1 \\ 2\cos^2(x)=\frac{2+\sqrt{2}}{2} \\ \cos^2(x)=\frac{2+\sqrt{2}}{4} \]
but yucky this looks a whole bunch uglier

Answer 21

ooo yeah definitely better the way you showed me.. thank you.. next one?

Answer 22

btw i like how you actually explain whats going on its greatly appreciated

Answer 23

take the square root of both sides
you will end solving two equations

Answer 24

\[\sin^2(x)=\frac{1}{2} \\ \text{ implies you have } \\ \sin(x)=\frac{1}{\sqrt{2}} \text{ or } \sin(x)=\frac{-1}{\sqrt{2}}\]
find the solutions to both equations
and the solution will be the union of the sets of solutions you found

Answer 25

45?

Answer 26

and -45?

Answer 27

earlier you said you wanted the solutions between 0 and 360
\[\sin(x)=\frac{\sqrt{2}}{2} \text{ has solutions } x=45^o \text{ or } x=? \\ \sin(x)=-\frac{\sqrt{2}}{2} \text{ has solutions } x=-45^o+360^o \text{ or } x=?\]
you still have 2 more solutions (1 for each equation)

Answer 28

315

Answer 29

well you already have the solution

Answer 30

that was the -45+360 one

Answer 31

i found 45, 315, 135 225

Answer 32

awesome sauce

Answer 33

number 59 is very similar to 57

Answer 34

number 60 is a quadratic
try solve 2u^2-1-u=0
( notice I just replaced cos(X) with u)

Answer 35

solve for u
you can use the quadratic formula if you want

Answer 36

then replace u with cos(x)
and then solve for x

Answer 37

i will be back in like 30 minutes (sorry)

Answer 38

OMGGGG I GET IT! THANK YOU SO MUCH!!!!!!!! (hope the pie tastes bomb!)

Answer 39

l messed up on the crust but the filling should taste great.

Answer 40

haha aww its okay.. tbh ive never had pecan pie.. but i have one last question

Answer 41

sure what is it

Answer 42

just curious how old are you bc youre INCREDIBLE at math

Answer 43

There is some math superior to me (a lot of math actually) but I'm 30.

Answer 44

ahh gotcha... that makes me feel ALOT better haha.. im only a senior in high school and if you were my age and that mathematically inclined id be awfully embarrasssed lol

Answer 45

I would write purely in terms of sin(x)
which will result in a quadratic equation again:
\[\cos(2x)=\cos^2(x)-\sin^2(x) \\ \cos(2x)=(1-\sin^2(x))-\sin^2(x) \\ \cos(2x)=1-2 \sin^2(x)\]
you would be embarrassed?

Answer 46

oops I mean in terms of cos(x)

Answer 47

one sec

Answer 48

yes i would bc compared to you im math illiterate...

Answer 49

\[\cos(2x)=\cos^2(x)-\sin^2(x) \\ \cos(2x)=\cos^2(x)-(1-\cos^2(x)) \\ \cos(2x)=2\cos^2(x)-1 \\ \text{ and } \sin^2(x)=1-\cos^2(x)\]
I forgot about that one term being cos(x) :p

Answer 50

\[\sin^2(x)+\cos(2x)-\cos(x)=0 \\ \text{ is equivalent to } \\ 1-\cos^2(x)+2\cos^2(x)-1-\cos(x)=0 \\ \text{ I replaced } \sin^2(x) \text{ with } 1-\cos^2(x) \\ \text{ and I replaced } \cos(2x) \text{ with } 2\cos^2(x)-1\]

Answer 51

\[1-\cos^2(x)+2\cos^2(x)-1-\cos(x)=0 \\ \text{ combine like terms on the \left } \\ \cos^2(x)-\cos(x)+1-1=0 \\ \cos^2(x)-\cos(x)=0\]
and actually this is tons easier to solve than using the quadratic formula
this one is easily factorable on the the left hend side

Answer 52

\[a^2-a=a(a-1)\]

\[\cos^2(x)-\cos(x)=\cos(x)(\cos(x)-1)\]
you will have two equations to solve
\[\cos(x)(\cos(x)-1)=0 \\ \text{ gives you } \cos(x)=0 \text{ or } \cos(x)-1=0\]

Answer 53

soooo 0 90 180

Answer 54

not 180
i meant 270

Answer 55

cos(x)=0 when x=90 or x=270 good job there
the other equation can be written as cos(x)=1

Answer 56

360

Answer 57

for when cos(x)=1

Answer 58

one question is the interval in which we solve the equation
(0,360)
or
[0,360)
or
(0,360]
or
[0,360]

Answer 59

[0,360)

Answer 60

so it would be 0 not 360

Answer 61

ok so [0,360)
means we don't look at 360 just everything up to it
and we do include 0 and things after you know until we get to 360 (which we do not include)

Answer 62

so instead of saying the solution to cos(x)=1 is 360 we will say 0

Answer 63

however if we were solving cos(x)=1 on [0,360] then we would say x=0 or x=360
since we can include both endpoints

Answer 64

yesss gotcha! seriously thank you soooo much :)

Answer 65

np

Answer 66

have a great night!

Answer 67

you too!