anonymous
  • anonymous
please help... FAN AND MEDAL
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
@freckles
freckles
  • freckles
which one do you looking at?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

freckles
  • freckles
all of them?
anonymous
  • anonymous
yes all
freckles
  • 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
  • anonymous
the interval is between 0 and 360
anonymous
  • anonymous
45 and 315
freckles
  • freckles
ok if we have \[0
freckles
  • freckles
now recall \[\theta=2x \\ \text{ so we have } \\ 2x=45^o ,315^o,405^o,675^o\]
freckles
  • freckles
just divide both sides by 2
anonymous
  • anonymous
so 22.2?
anonymous
  • anonymous
157.5?
freckles
  • freckles
ok and two more solutions
freckles
  • freckles
\[x=\frac{45^o}{2},\frac{315}{2}^o,\frac{405^o}{2},\frac{675^o}{2}\]
freckles
  • freckles
we had to solve cos(theta)=sqrt(2)/2 in (0,720) since x was between 0 and 360
freckles
  • freckles
theta was 2 times more than x
anonymous
  • anonymous
ahh gotcha.. and thats the answer?
freckles
  • freckles
yes those 4 solutions right there at mentioned are the answers
anonymous
  • anonymous
is that considered double identity angles?
freckles
  • 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
  • anonymous
ooo yeah definitely better the way you showed me.. thank you.. next one?
anonymous
  • anonymous
btw i like how you actually explain whats going on its greatly appreciated
freckles
  • freckles
take the square root of both sides you will end solving two equations
freckles
  • 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
  • anonymous
45?
anonymous
  • anonymous
and -45?
freckles
  • 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
  • anonymous
315
freckles
  • freckles
well you already have the solution
freckles
  • freckles
that was the -45+360 one
anonymous
  • anonymous
i found 45, 315, 135 225
freckles
  • freckles
awesome sauce
freckles
  • freckles
number 59 is very similar to 57
freckles
  • freckles
number 60 is a quadratic try solve 2u^2-1-u=0 ( notice I just replaced cos(X) with u)
freckles
  • freckles
solve for u you can use the quadratic formula if you want
freckles
  • freckles
then replace u with cos(x) and then solve for x
freckles
  • freckles
i will be back in like 30 minutes (sorry)
anonymous
  • anonymous
OMGGGG I GET IT! THANK YOU SO MUCH!!!!!!!! (hope the pie tastes bomb!)
freckles
  • freckles
l messed up on the crust but the filling should taste great.
anonymous
  • anonymous
haha aww its okay.. tbh ive never had pecan pie.. but i have one last question
freckles
  • freckles
sure what is it
anonymous
  • anonymous
just curious how old are you bc youre INCREDIBLE at math
freckles
  • freckles
There is some math superior to me (a lot of math actually) but I'm 30.
anonymous
  • anonymous
1 Attachment
anonymous
  • 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
  • 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
  • freckles
oops I mean in terms of cos(x)
freckles
  • freckles
one sec
anonymous
  • anonymous
yes i would bc compared to you im math illiterate...
freckles
  • 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
  • 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
  • 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
  • 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
  • anonymous
soooo 0 90 180
anonymous
  • anonymous
not 180 i meant 270
freckles
  • freckles
cos(x)=0 when x=90 or x=270 good job there the other equation can be written as cos(x)=1
anonymous
  • anonymous
360
anonymous
  • anonymous
for when cos(x)=1
freckles
  • freckles
one question is the interval in which we solve the equation (0,360) or [0,360) or (0,360] or [0,360]
anonymous
  • anonymous
[0,360)
anonymous
  • anonymous
so it would be 0 not 360
freckles
  • 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
  • freckles
so instead of saying the solution to cos(x)=1 is 360 we will say 0
freckles
  • 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
  • anonymous
yesss gotcha! seriously thank you soooo much :)
freckles
  • freckles
np
anonymous
  • anonymous
have a great night!
freckles
  • freckles
you too!

Looking for something else?

Not the answer you are looking for? Search for more explanations.