please help... FAN AND MEDAL

- anonymous

please help... FAN AND MEDAL

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- schrodinger

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- anonymous

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- anonymous

@freckles

- freckles

which one do you looking at?

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## More answers

- freckles

all of them?

- anonymous

yes all

- freckles

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 ?

- anonymous

the interval is between 0 and 360

- anonymous

45 and 315

- freckles

ok if we have
\[0

- freckles

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

- freckles

just divide both sides by 2

- anonymous

so 22.2?

- anonymous

157.5?

- freckles

ok and two more solutions

- freckles

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

- freckles

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

- freckles

theta was 2 times more than x

- anonymous

ahh gotcha.. and thats the answer?

- freckles

yes those 4 solutions right there at mentioned are the answers

- anonymous

is that considered double identity angles?

- freckles

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

- anonymous

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

- anonymous

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

- freckles

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

- freckles

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

- anonymous

45?

- anonymous

and -45?

- freckles

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)

- anonymous

315

- freckles

well you already have the solution

- freckles

that was the -45+360 one

- anonymous

i found 45, 315, 135 225

- freckles

awesome sauce

- freckles

number 59 is very similar to 57

- freckles

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

- freckles

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

- freckles

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

- freckles

i will be back in like 30 minutes (sorry)

- anonymous

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

- freckles

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

- anonymous

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

- freckles

sure what is it

- anonymous

just curious how old are you bc youre INCREDIBLE at math

- freckles

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

- anonymous

##### 1 Attachment

- anonymous

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

- freckles

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?

- freckles

oops I mean in terms of cos(x)

- freckles

one sec

- anonymous

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

- freckles

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

- freckles

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

- freckles

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

- freckles

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

- anonymous

soooo 0 90 180

- anonymous

not 180
i meant 270

- freckles

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

- anonymous

360

- anonymous

for when cos(x)=1

- freckles

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

- anonymous

[0,360)

- anonymous

so it would be 0 not 360

- freckles

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)

- freckles

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

- freckles

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

- anonymous

yesss gotcha! seriously thank you soooo much :)

- freckles

np

- anonymous

have a great night!

- freckles

you too!

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