Question

HELP! MEDAL! @SolomonZelman @texaschic101 @Abhisar @dan815 @wio @Kainui @TheSmartOne

Answer 1

re-write \(\large\color{slate}{\sin^2(x) }\) in terms of \(\large\color{slate}{ \cos(x) }\)

using: \(\large\color{slate}{ \sin^2x+\cos^2x=1 }\)

then, say \(\large\color{slate}{ {\rm let}~~~\cos(x)=a }\) and solve the quadratic.

Answer 2

Okay, hold on a sec while I do that...

Answer 3

\(\large 4[1-cos^2(x)]+4\sqrt{2}[cos(x)]-6=0\)
\(\large 4-4[cos^2(x)]+4\sqrt{2}[cos(x)]-6=0\)
\(\large -4a^2+4\sqrt{2}a-2=0\)
Like that?

Answer 4

yes, like this.

Answer 5

okay, then use quadratic formula?

Answer 6

put \ before cos(x)

Answer 7

no, I would complete the square:

Answer 8

oh, okay

Answer 9

\(\large\color{slate}{ (\sqrt{2}/2)^2=2/4=1/2 }\)

Answer 10

It would be \(4\sqrt{2}\) not \(\sqrt{2}\)

Answer 11

\(\Large(\frac{4\sqrt{2}}{2})^2\rightarrow\frac{16(2)}{4}\rightarrow4(2)=8\)
Rigth?

Answer 12

Right?

Answer 13

\(\large\color{slate}{-4a^2+4\sqrt{2}a-2=0 }\)

\(\large\color{slate}{-4a^2+4\sqrt{2}a=2}\)

\(\large\color{slate}{-4(a^2-\sqrt{2}a)=2}\)

\(\large\color{slate}{a^2-\sqrt{2}a=-1/2}\)

\(\large\color{slate}{a^2-\sqrt{2}a+1/2=-1/2+1/2}\)

Answer 14

is this making sense?

Answer 15

Yeah, that makes sense. :)

Answer 16

finish completing the square

Answer 17

Not sure how I factor it though...
\(\large a^2-\sqrt{2}a+1/2=0\)

Answer 18

\(\large\color{slate}{(\sqrt{2}/2)^2=2/4=1/2}\)

Answer 19

perf. sq. trinomial

Answer 20

so...
\(\large (a+\sqrt{2}{2})^2=0\)
?

Answer 21

oops, forgot the \

Answer 22

you mean \(\large\color{slate}{\sqrt{2}/2}\) ?

Answer 23

right

Answer 24

I see, you get it though...

Answer 25

for some reason teh "\" won't show up

Answer 26

and it is negative

Answer 27

oh, right.

Answer 28

\(\large\color{slate}{a^2-\sqrt{2}a+1/2=0}\)

\(\large\color{slate}{(a-\sqrt{2}/2)^2=0}\)

Answer 29

yeah, that, lol ^

Answer 30

okay, solve for a, now

Answer 31

So then \(\large a=\Large\frac{\sqrt{2}}{2}\)
right?

Answer 32

yes

Answer 33

it's a double root i think

Answer 34

Okay, so what's the next step?

Answer 35

your next step is, to sub back...
you had, \(\large\color{slate}{a=\cos(x)}\) remember?

Answer 36

right
OH I SEE

Answer 37

so \(\large\color{slate}{\cos(x)=\frac{\LARGE \sqrt{2} }{\LARGE 2}}\)

Answer 38

x = pi/4 then

Answer 39

Yes, it is a 45-45-90 triangle:)
45deg, or pi/4

Answer 40

and 7pi/4
So B.
THANK YOU SO VERY MUCH. My textbook didn't have that in it

Answer 41

\(\large\color{slate}{ \rm~Very~good! }\)

Answer 42

Thanks again. You're awesome. :)

Answer 43

You get even smallest tips, and have a very very appropriate background knowledge!! tY!

Answer 44

B is right

Answer 45

I wish I was back in elementary school where math was easy-peasy. Now I'm in 10th grade, and math is HARD