Question

Solve for x: \(\sin{\frac{x}{2}} = 2\cos^{2}x - 1\)
Please guide me to the answer. Like a push in the right direction.

Answer 1

sin x/2 might be better expressed in sin cos form

Answer 2

Like:\[\sqrt{\frac12(1 - \cos(x))} = 2\cos^2x - 1\]?

Answer 3

Made a slight typo:
\[\sin^2\left(\frac{x}{2}\right)=\frac{1}{2}-\frac{1}{2}\cos x\]

Answer 4

How did you get there?

Answer 5

It's an identity. Half-angle is the name. You have the same thing as your first step, it seems.

Answer 6

Ah I see. You removed the √ and distributed.

Answer 7

sin^2 + cos^2 = 1

2cos^2 - (cos^2 + sin^2)

cos^2 x - sin^2 x= cos(2x)

Answer 8

While trying to work this problem, I just end up going in circles...I can't seem to get anywhere no matter what identities I use.

Answer 9

do you have to use identities?

Answer 10

I think that's how the problem is designed, but I suppose other means would work as well. As long it doesn't involve a calculator or a graphing utility.

Answer 11

http://www.wolframalpha.com/input/?i=sin%28x%2F2%29+%3D+2cos%28x%29-1

doesnt seem like there will be a nice outcome to it ....

Answer 12

forgot to ^2

Answer 13

yeah, thats still messy

Answer 14

you sure youve got it posted correctly?

Answer 15

Yes. This is exactly what is on the sheet.

Answer 16

hmm

cos(2x) = sin(x/2) ; let x = 2u

cos(4u) = sin(u) might be a way to get rid of those fractions to see things better

Answer 17

Right now, I'm trying to get all of them to only one trig function so that I can solve it like a polynomial. I did manage to eliminate the different angles, but it's really messy.

Answer 18

looks like we can get something like:\[sin(u)=sin^4(u)+cos^4(u)-6sin^2(u)cos^2(u)\]

Answer 19

as far as getting that into a poly; its hideous as well

Answer 20

With what you gave me, I simplified to:\[1 - \sin(u) = 8\sin^2u\cos^2u\]

Answer 21

i think you misplaced a negative with that .... im getting a mirrored graph of the 2

Answer 22

I'm fairly certain it is correct. I used \((\sin^2(u) + \cos^2(u))^2\) to get \(1 = \sin^4(u) + \cos^4(u) + 2\sin^2(u)\cos^2(u)\) and subtracted the two equations.

Answer 23

just seeing this is all ....

http://www.wolframalpha.com/input/?i=y+%3D+sin%28x%29-cos%284x%29%2C+y+%3D+1-sin%28x%29-8sin%5E2%28x%29cos%5E2%28x%29

Answer 24

prolly how i subtracted the first one ... sholda went the same way with both

Answer 25

thats better :)

Answer 26

cos^2 = 1 - sin^2

1 - sin = 8 sin^2(1-sin^2)

1 - sin = 8 sin^2 - 8sin^4

8sin^4 -8sin^2 -sin +1 = 0

Answer 27

Yup. I'm factoring it to see what I can get.

Answer 28

One of the roots is π.

Answer 29

just noticed a difference of squares ...

1 - sin = 8 sin^2(1-sin^2)

1 - sin = 8 sin^2(1-sin)(1+sin)

1 = 8 sin^2(1+sin)

8 sin^2+8sin^3 - 1 = 0

Answer 30

Hmm...This is what I have at the moment.\[(\sin(u) - 1)[(8\sin^2(u))(\sin(u) + 1) - 1] = 0\]That's how I got the x = π.

Answer 31

Which is what you have in the second grouping.

Answer 32

yep

Answer 33

there are 4 primary results ...

Answer 34

Well, thanks for the help anyway. We did manage to find one of the roots. I'm sure that I'll figure it out if I play with it enough.

Answer 35

my concern is that they are not your basic "no calculator or graphing" run of the mill solutions ...

Answer 36

7pi/20 doesnt seem like a standard angle to me

Answer 37

Yeah. This isn't a typical class problem. It is a problem on a math team test, so it would normally be done under more stringent solutions. I do have the answers, but I am trying not use them. Apparently, the two remaining roots were: x = π/3 and 5π/3.

Answer 38

:) good luck with it all

Answer 39

there were 4 primary solutions of the original setup

Answer 40

Thank you :)