Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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

Mathematics
See more answers at brainly.com
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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

120,360
oh my, beaten to the punch gj

Not the answer you are looking for?

Search for more explanations.

Ask your own question

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
yes!
Using Q? Oooo I like that :) I'm always having trouble finding a suitable letter for theta in text, that works quite well XD
^concur
lol ya, i got tired ofputting theata.
30
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.
yup^
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
lol,
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
uh
haha did i do it right?!
yes
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

Not the answer you are looking for?

Search for more explanations.

Ask your own question