Babynini
  • Babynini
Solve each equation in the interval [0,2pi) rounded to two decimal points a) (2cos(theta)-1)(sin(theta)-1)=0
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Babynini
  • Babynini
em I got 2cos(theta)sin(theta)-2cos(theta)-sin(theta)+1=0 yeah?
freckles
  • freckles
If a*b=0, then either a=0 or b=0 or both=0. This means you have: \[2\cos(\theta)-1=0 \text{ or } \sin(\theta)-1=0\]
freckles
  • freckles
isolate the trig function

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

freckles
  • freckles
\[\cos(\theta)=\frac{1}{2} \text{ or } \sin(\theta)=1 \]
freckles
  • freckles
try to solve both of those equations
Babynini
  • Babynini
cos(theta) = 1.047 sin(theta) = 1.57
freckles
  • freckles
oh you mean: \[\theta \approx 1.05 \text{ you should also be able to obtain } \theta \approx 5.24 \text{ from } \cos(\theta)=\frac{1}{2}\] and \[\theta \approx 1.57 \text{ from the other equation } \sin(\theta)=1 \]
Babynini
  • Babynini
Yes, that is what I meant :) how did you get 5.24?
freckles
  • freckles
\[\cos(\theta)=\cos(-\theta)=\cos(-\theta+2 \pi ) \\ \cos(1.05) \approx \cos(-1.05+2\pi) \approx \cos(-1.05+2(3.14)) \\ \cos(1.05) \approx \cos(-1.05+6.28)=\cos(5.23) \\ \text{ but honestly I used the unit circle }\]
freckles
  • freckles
https://upload.wikimedia.org/wikipedia/commons/thumb/4/4c/Unit_circle_angles_color.svg/720px-Unit_circle_angles_color.svg.png
freckles
  • freckles
You see the points on the circle. The first number of every pair is the output for the cos(the angle there). The second number of every pair is the output for the sin(the angle there). I was looking for when the x-coordinate (the first number aka the cos number) was 1/2. I see this was happening at the pi/3 mark and the 5pi/3 mark.
freckles
  • freckles
And your question wanted these values rounded to the nearest hundredth.
freckles
  • freckles
\[\frac{\pi}{3} \approx 1.05 \\ \frac{5 \pi}{3} \approx 5.24\]
Babynini
  • Babynini
ooo I see! that makes sense! haha why do we not want to do the othe pi/3 s?
freckles
  • freckles
at 2pi/3 and 4pi/3 you should see the cos number (the x-coordinate) is negative
freckles
  • freckles
1/2 is not a negative number
Babynini
  • Babynini
at pi/3 it is not negative either. That was another question I have why do we make it -1.05 instead of adding 2pi to 1.05
freckles
  • freckles
cos is an even function \[\cos(\theta)=\cos(-\theta)\]
freckles
  • freckles
sin is an odd function \[\sin(\theta)=-\sin(-\theta)\]
Babynini
  • Babynini
ooh yes.
freckles
  • freckles
and then inside you can add as many 2pi as you like because we just wind up at the same place
freckles
  • freckles
but I add one 2pi since that would give me another solution in [0,2pi)
freckles
  • freckles
Pretend we want to solve this: \[\cos(\theta)=.23 , \theta \in [0,2\pi)\] This .23 is not on the unit circle. But using the unit circle and other identities helps me to solve this equation. First I can do arccos( ) on both sides to find one solution. \[\theta=\arccos(.23) \] another solution would be given by: \[-\theta+2\pi=\arccos(.23) \\ \theta=-\arccos(.23)+2\pi\] and if we wanted to approximate these solutions by rounding to the nearest hundredth we would have: \[\theta \approx 4.94 , 1.34 \]
freckles
  • freckles
The second solution I used the fact that cos was even and had a period of 2pi.
Babynini
  • Babynini
Hmm ok.
Babynini
  • Babynini
Babynini
  • Babynini
Thank you for the help on this one, btw. :)
freckles
  • freckles
np

Looking for something else?

Not the answer you are looking for? Search for more explanations.