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Determine all of the solutions in the interval 0< Theta < 360. cos(Q/2) = 1/2 Q is theta

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oh my, beaten to the punch gj

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Other answers:

haha okay so then hmmm
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?
well, now that I think about it, I think raja might be wrong
it should have 4 solutions right?
im not sure, it just says to determine all solutions in the interval 0-->360
Using Q? Oooo I like that :) I'm always having trouble finding a suitable letter for theta in text, that works quite well XD
lol ya, i got tired ofputting theata.
okay is it 60 and 300 degrees?
no tetha is 30
\[\large \cos(\theta/2)=\frac{1}{2}\]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?
look, as I solve it, it's \[\theta=2\cos^{-1} (1/2)\]
the thing is, that only gives you 120. but it's a good starting point
mmmmm... okay, well first. special angles is like 45-45-90, and 30-60-90, right?
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.
oh....okay i gotcha now on special angles. lol im totaly flipped around in this class
okay then so..cosQ/2 = 1/2 right? so cos would be a positive
I think 120 is it.....
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.
yes :) 1st and 4th
so, if we find the reference angle, then what would the two angles be?
mmm....pi/3 and 5pi/6?
yes to the first, the last, not so much, that's in the second quadrant
or am I failing?
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 :)
oh. haha i mean 5pi/3
With zepdrix, but you are right
so the second answer would be 5pi/3?
but, the problem is, we need that angle to be doubled to match our equation of theta/2
hmmm...ok okkay, so doubled from pi/3, so pi/3 and 5pi/6?
uh, lets throw this back to degrees shall we? pi/3 = 60 and 5pi/3 = 300
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
no, 5pi/6 = 150 degrees
haha shoot what am i doing wrong
here's a tip. every pi/6 = 30 degrees
pi/6 pi/4 pi/3 pi/2 4pi/6div by 2 = 2pi/3 OH!.....haha okay so pi/3 and 2 pi/3???
just plug those into your original equation and see what you get :P
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?
@zepdrix is it pi/3 and 2pi/3?
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.
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
AH! so i was right?! the 60 and 300?!
\[\large \theta=\quad 120^o, \quad 600^o\]
and since we're going only to 360?
(radians are easier imo but degrees work k)
AH!!! im so lost haha
aw :D
its almost like you're my lilsis or somethin xD
okay so i get the 1/2 and that is in the first quad giving me 60 degrees right?
yes, and the 300
okay, how can i figure out the 300?!
it's a reference angle of 60
yes it is a ref angle 60
WAIT!!! 300 cuz its cos 1.2, sin - sqrt3 /2 ?!
right?! dang i think im slow
haha did i do it right?!
YAY!!!! > . < THANK YOU SO MUCH!!! ugh....
however, you have to double those and since you only go to 360, 120 is your only answer
  • phi
sometimes sketching the curve is helpful |dw:1355239813847:dw|
agreed, but my calculator is dead xD

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