Babynini
  • Babynini
5cos2(theta)=2 Find all solutions in interval [0,2pi)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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Babynini
  • Babynini
so cos2(theta)=2/5 2(theta)= ?
Babynini
  • Babynini
@peachpi it's a little different.. :P
anonymous
  • anonymous
Is that cos 2Θ or cos² Θ?

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Babynini
  • Babynini
the first
anonymous
  • anonymous
ok. so the first thing I like to do when Θ is multiplied is "fix" the interval. 0 ≤ x < 2π Since the x is multiplied by 2, multiply the whole interval by 2 to get 0 ≤ 2x < 4π
anonymous
  • anonymous
so we do the same thing we did last time, except now we're looking for solutions from 0 to 4pi. cos 2Θ = 2/5 2Θ = arccos (2/5)
Babynini
  • Babynini
arccos(2/5) = 1.159 so we want to add pi to this?
Babynini
  • Babynini
or, sorry, subtract this from 2pi
anonymous
  • anonymous
You want to divide that by 2 Θ = ½ arccos (2/5) → answer in 1st quadrant on the interval from 0 to 2pi Then add 2pi to get the 1st quad angle between 2pi and 4pi
Babynini
  • Babynini
em so the original answer is 0.579 adding 2pi = 6.862
anonymous
  • anonymous
the 0.579 is right. and you actually subtract it from 2pi to get the other answer
anonymous
  • anonymous
so 5.7
Babynini
  • Babynini
*5.8 :)
anonymous
  • anonymous
Yes. 5.8 :) There are going to be two more solutions. You got 2Θ = 1.159. Add 2pi to that, then divide by 2 to get the corresponding Θ.
anonymous
  • anonymous
should get 3.72
Babynini
  • Babynini
Yep yep
anonymous
  • anonymous
and then (2pi - 1.159)/2 for the 4th solution is 2.56
Babynini
  • Babynini
K so it's theta = 1.159/2 = 0.579 So our first answer: 5.8 add 2pi to that: 3.72 next we have 2pi-0.58 = 5.70 (our third answer) then 5.70 - pi = 2.56 (our fourth answer) haha sorry that's all jumbled up.
Babynini
  • Babynini
= 0.58, 3.72,5.70,2.56
anonymous
  • anonymous
yep
Babynini
  • Babynini
I'll get this. Sometime. thanks!
anonymous
  • anonymous
you're welcome
Babynini
  • Babynini
@peachpi I just did one by myself!! and got the right answer! xD :')
anonymous
  • anonymous
awesome!

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