A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

cassieforlife5

  • one year ago

Solve the equation in the following intervals: sin x = -0.2 I know for this you have to find sin^-1 (-0.2) and I got -0.201 (my teacher said we have to do it in radians not degrees) but i'm so confused from here! Can anyone explain how to solve it?

  • This Question is Closed
  1. cassieforlife5
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    The intervals are: \[0 \le x < 2\pi \]

  2. cassieforlife5
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    and the second one is: \[-\infty < x < \infty \]

  3. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[x=-0.201358+2n \Pi\] for \[0 \le x < 2\pi\]: x=−0.201358+2Π for \[-\infty < x < \infty\]: x=−0.201358+2nΠ, \[n=0,\pm1,\pm2,\pm3,....\]

  4. cassieforlife5
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Thanks for your reply! Can you explain why you add 2pi for the 1st one and 2npi for the second one?

  5. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    because the period of sin x is 2Pi, it repeat it self after 2pi.

    1 Attachment
  6. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    for 0≤x<2π : x=−0.201358+2Π this because the interval, −0.201358<0 so this solution is rejected x=−0.201358+2Π is good x=−0.201358+2*2Π>2Π this solution is rejected

  7. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Do you get it?

  8. cassieforlife5
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh i see. so i have another question if you can help?

  9. cassieforlife5
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    sec x= -3 for the interval \[-\pi \le x < \pi \] 1/cosx=-3 cosx= -1/3 cos^-1 (-1/3)= 1.911 + pi is this the only answer or are there more?

  10. cassieforlife5
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okay so 1.911-pi is the only answer

  11. cassieforlife5
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yeah sorry forgot to include that part but i got it

  12. cassieforlife5
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    gosh you're a life saver!! could you check my work for this one as well?

  13. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    sorry I made mistake

  14. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    x=1.911

  15. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    this one solution

  16. cassieforlife5
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    cot x= -1 for -infinity< x < infinity cosx/sinx= -1 3pi/4+ 2pi and 7pi/4+ 2pi ?

  17. cassieforlife5
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    sorry i meant 3pi/4+ npi 7pi/4 + 4pi

  18. cassieforlife5
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    7pi/4+ npi

  19. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    for the previous problem the solution is x=1.911

  20. cassieforlife5
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yeah because you don't subtract possibilities right?

  21. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    for the last on :x= 3pi/4+ 2npi and 7pi/4+ 2npi

  22. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    *one

  23. cassieforlife5
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    is it 2npi? i thought it was just npi because the period of cotangent is pi?

  24. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    cot x=cos x/sin x, so it is contained on two functions, and period of cos x and sin x is 2npi

  25. cassieforlife5
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh... alright. sorry for asking so many questions but honestly you made it so much easier!! 2sin(2theta)+\[\sqrt{3}\]= 0 [0, 2pi) I got 4pi/3 and 5pi/3. It says to give both general and specific solutions, but these were the only ones that fit the intervals

  26. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[2\sin (2\theta)+ \sqrt{3}=0?\]

  27. cassieforlife5
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yeah sorry it got typed awkwardly

  28. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    let solve it

  29. cassieforlife5
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    2sin2x= - sqrt(3) sin2x= - sqrt(3)/2 at 4pi/3 and 5pi/3 I've added 2npi, but all the solutions were larger than 2pi so I thought they were invalid

  30. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I got (-pi/6)+npi, (-pi/3)+npi

  31. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    this in general without considering the interval

  32. cassieforlife5
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    could you explain? i think it might be because I didn't get rid of the 2 in sin2x= -sqrt(3)/2

  33. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    your answer is also correct.

  34. cassieforlife5
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    which one is correct?

  35. cassieforlife5
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    is this right? sin2x= \[-\sqrt{3}/2 \] 2x= 4pi/3 or 2x= 5pi/3 x= 2pi/3 or 5pi/6

  36. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    correct and there are two more solutions.

  37. cassieforlife5
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    2 more? I have no idea how to get them then

  38. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    because npi

  39. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    and the interval [0, 2pi)

  40. cassieforlife5
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I thought it was supposed to be 2npi? and i tried that and it went over 2pi

  41. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    2x= 4pi/3 +2npi, 5pi/3+2npi x= 2pi/3 +npi, 5pi/6+npi

  42. cassieforlife5
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Oh I didn't know you added the 2npi before simplifying

  43. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I forget it, you should and it

  44. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    for the interval [0, 2pi): \[x=\frac{ 2\pi }{ 3 } , \frac{ 2\pi }{ 3 }+\pi=\frac{ 5\pi }{ 3 },\frac{ 5\pi }{ 6 } ,\frac{ 5\pi }{ 6 }+\pi=\frac{ 11\pi }{ 6}\]

  45. cassieforlife5
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yup I checked all 4 answers and they all equalled 0!! 2sin^2x= 2+cosx [0,pi] after using an identity and simplifying, I got cosx(2cosx +1)=0 cosx=0 at pi/2 and 3pi/2 2cosx +1-> cosx= -1/2 at 2pi/3 and 4pi/3 I eliminated 3pi/2 and 4pi/3 because they were bigger than pi so my answers are pi/2 and 2pi/3

  46. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I need to eat my dinner. can we complete it later

  47. cassieforlife5
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    sure! thank you so much for your help!!! i'll work on other things until then

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

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.