Question

Determine all of the solutions in the interval 0< Theta < 360.

cos(Q/2) = 1/2

Q is theta

Answer 1

@zepdrix

Answer 2

120,360

Answer 3

oh my, beaten to the punch gj

Answer 4

haha okay so then hmmm

Answer 5

how would that be? @zepdrix i dont get it. i see cos Q/2 that would mean cos = 1/2 right? and if its positive it would be in the 1st and 4th Quadrant?

Answer 6

well, now that I think about it, I think raja might be wrong

Answer 7

it should have 4 solutions right?

Answer 8

im not sure, it just says to determine all solutions in the interval 0-->360

Answer 9

yes!

Answer 10

Using Q? Oooo I like that :) I'm always having trouble finding a suitable letter for theta in text, that works quite well XD

Answer 11

^concur

Answer 12

lol ya, i got tired ofputting theata.

Answer 13

30

Answer 14

okay is it 60 and 300 degrees?

Answer 15

no tetha is 30

Answer 16

\[\large \cos(\theta/2)=\frac{1}{2}\]We need to solve for theta, which involves a little more than just finding a special angle. If it helps, think of the ENTIRE inside of cosine as an angle, like this,\[\large \cos(\phi)=\frac{1}{2}\]Where is phi =1/2? Which special angles?

Answer 17

look, as I solve it, it's
\[\theta=2\cos^{-1} (1/2)\]

Answer 18

the thing is, that only gives you 120. but it's a good starting point

Answer 19

mmmmm... okay, well first. special angles is like 45-45-90, and 30-60-90, right?

Answer 20

The special angles are 0,30, 45, 60, 90
In radians 0, pi/6, pi/4, pi/3, pi/2 and so on...

Remember those? :)
We're not talking about special triangles :D
Special angles on the unit circle.

Answer 21

oh....okay i gotcha now on special angles. lol im totaly flipped around in this class

Answer 22

okay then so..cosQ/2 = 1/2 right?
so cos would be a positive

Answer 23

I think 120 is it.....

Answer 24

yes the cosine is producing a positive value, so we're concerned with angles in the 1st and 4th quadrant right? :O
Cuz that's where cosine is positive.

Answer 25

yes :) 1st and 4th

Answer 26

so, if we find the reference angle, then what would the two angles be?

Answer 27

mmm....pi/3 and 5pi/6?

Answer 28

yes to the first, the last, not so much, that's in the second quadrant

Answer 29

or am I failing?

Answer 30

lol :) They should both either be /6 or /3 :D you should get similar angles in that sense.

Also, the domain of theta was given to us in DEGREES. let's try to keep it in degrees :)

Answer 31

oh. haha i mean 5pi/3

Answer 32

With zepdrix, but you are right

Answer 33

so the second answer would be 5pi/3?

Answer 34

but, the problem is, we need that angle to be doubled to match our equation of theta/2

Answer 35

hmmm...ok okkay, so doubled from pi/3, so pi/3 and 5pi/6?

Answer 36

uh, lets throw this back to degrees shall we?

pi/3 = 60 and 5pi/3 = 300

Answer 37

DANG! haha okay, so first we have 60 and u said to double it, so that would be 120, 120 would be.....that 5pi/6

Answer 38

no, 5pi/6 = 150 degrees

Answer 39

haha shoot what am i doing wrong

Answer 40

here's a tip. every pi/6 = 30 degrees

Answer 41

pi/6 pi/4 pi/3 pi/2 4pi/6div by 2 = 2pi/3

OH!.....haha okay so pi/3 and 2 pi/3???

Answer 42

just plug those into your original equation and see what you get :P

Answer 43

i dont have an equatioN haha says determine all of the solutions in the interval 0< Q<360

cos(Q/2) = 1/2

so the answers would be pi/3 and 2pi/3? right?

Answer 44

@anikay

Answer 45

@zepdrix is it pi/3 and 2pi/3?

Answer 46

If the problem had been,\[\large \cos\left(\theta\right)=\frac{1}{2}\]We would have concluded that,\[\large \theta=\quad 60^o,\quad 300^o\]

But our problem involved Q/2, not Q.
Giving us,\[\large \cos\left(\frac{\theta}{2}\right)=\frac{1}{2}\]
\[\frac{\theta}{2}=\quad 60^o,\quad 300^o\]
From here, if we multiply both sides by 2, we can solve for Q.

Answer 47

yup^

Answer 48

I'm a little confused why you're messing with the radians :C They're a little more confusing to work with.

The domain of Q was GIVEN to us in DEGREES. So let's work in degrees! :D

Answer 49

AH! so i was right?! the 60 and 300?!

Answer 50

\[\large \theta=\quad 120^o, \quad 600^o\]

Answer 51

and since we're going only to 360?

Answer 52

(radians are easier imo but degrees work k)

Answer 53

AH!!! im so lost haha

Answer 54

aw :D

Answer 55

its almost like you're my lilsis or somethin xD

Answer 56

lol,

Answer 57

okay so i get the 1/2 and that is in the first quad giving me 60 degrees right?

Answer 58

@anikay

Answer 59

yes, and the 300

Answer 60

okay, how can i figure out the 300?!

Answer 61

it's a reference angle of 60

Answer 62

yes it is a ref angle 60

Answer 63

WAIT!!! 300 cuz its cos 1.2, sin - sqrt3 /2 ?!

Answer 64

right?! dang i think im slow

Answer 65

uh

Answer 66

haha did i do it right?!

Answer 67

yes

Answer 68

YAY!!!! > . < THANK YOU SO MUCH!!! ugh....

Answer 69

however, you have to double those and since you only go to 360, 120 is your only answer

Answer 70

sometimes sketching the curve is helpful