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MTALHAHASSAN2

  • one year ago

Need help!! 8. Solve for theta. 0<theta<2pie. Given an exact answer, where possible. Otherwise, round to the nearest hundredth of a radians. 5 sin theta cos theta – 3 cos theta = 0

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  1. campbell_st
    • one year ago
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    factor out cos \[\cos(\theta)(5\sin(\theta) -3) \] now set each factor to zero and solve \[\cos(tehta) = 0~~~and~~~5\sin(\theta) - 3 = 0\]

  2. MTALHAHASSAN2
    • one year ago
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    |dw:1434604876572:dw|

  3. MTALHAHASSAN2
    • one year ago
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    can someone plz help me

  4. MTALHAHASSAN2
    • one year ago
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    @Michele_Laino

  5. MTALHAHASSAN2
    • one year ago
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    @KyanTheDoodle

  6. MTALHAHASSAN2
    • one year ago
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    @RezForLife

  7. MTALHAHASSAN2
    • one year ago
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    @Alexander95

  8. MTALHAHASSAN2
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    @bleebleebleed

  9. MTALHAHASSAN2
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    @Compassionate

  10. MTALHAHASSAN2
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  11. MTALHAHASSAN2
    • one year ago
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    @esam2

  12. MTALHAHASSAN2
    • one year ago
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    @freckles

  13. anonymous
    • one year ago
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    From your unit circle you can pick out where cosine is 0. For the sine solutions, you already have sin Θ = 0.6. Use the sin^-1 button on your calculator to find Θ Θ = sin^-1 (0.6) This gives an angle in the 1st quadrant. sine is also positive in the 2nd quadrant. Subtract the other solution from pi to get the angle in the 2nd quadrant. Θ = π - sin^-1 (0.6)

  14. anonymous
    • one year ago
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    do still need help i would have messaged u but i got blocked for god knows what

  15. anonymous
    • one year ago
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    @MTALHAHASSAN2

  16. MTALHAHASSAN2
    • one year ago
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    Θ = sin^-1 (0.6) i get 37 degree

  17. anonymous
    • one year ago
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    ican help or fleather me

  18. MTALHAHASSAN2
    • one year ago
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    @SkaterBoyShawn what is ur problem

  19. MTALHAHASSAN2
    • one year ago
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    I don't need ur help

  20. MTALHAHASSAN2
    • one year ago
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    @Michele_Laino

  21. anonymous
    • one year ago
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    i have no prob u said fleather me then blocked me i am confuesd make sense i dont know what i did to u

  22. UsukiDoll
    • one year ago
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    inb4 crazy ban from original poster ...stop mass tagging people.

  23. MTALHAHASSAN2
    • one year ago
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    @SkaterBoyShawn stop

  24. anonymous
    • one year ago
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    yes 37° is one solution. 180° - 37° = 143° is another solution

  25. anonymous
    • one year ago
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    now what did i do

  26. MTALHAHASSAN2
    • one year ago
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    wait but at the back the answer is different

  27. MTALHAHASSAN2
    • one year ago
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    it is something like 74

  28. anonymous
    • one year ago
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    stop tagging me iam bothing u so leave me alone bye have a good evning

  29. Michele_Laino
    • one year ago
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    we can factor out cos( \theta), so we can write: \[\Large \cos \theta \left( {5\sin \theta - 3} \right) = 0\]

  30. campbell_st
    • one year ago
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    you have missed on of the key aspects of the question, you are working in radians and you should have 4 angles as the solution

  31. MTALHAHASSAN2
    • one year ago
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    no@campbell

  32. MTALHAHASSAN2
    • one year ago
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    the answer is given in degrees

  33. anonymous
    • one year ago
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    @MTALHAHASSAN2 yes he's right. It is in radians

  34. campbell_st
    • one year ago
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    you solutions to the factor are correct.. \[\cos(\theta) = 0\] what 2 angles measure make this true, remember its in radians...

  35. anonymous
    • one year ago
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    "0<theta<2pie" is what you typed above

  36. MTALHAHASSAN2
    • one year ago
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    oh yeah

  37. MTALHAHASSAN2
    • one year ago
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    my bad

  38. campbell_st
    • one year ago
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    the domain is given as \[0 < \theta l< 2\pi\] so you should answer in radians

  39. MTALHAHASSAN2
    • one year ago
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    yep the answer is in radian

  40. campbell_st
    • one year ago
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    so the solutions for \[\cos(\theta) = 0\] should be straight forward the 2 angles for \[\sin(\theta) = \frac{3}{5}\] are in the 1st quadrant and 2nd quadrant

  41. imqwerty
    • one year ago
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    5sin(θ)cos(θ)-3cos(θ) = 0 divide by 5 sin(θ)cos(θ) -3/5cos(θ)=0 3/5=sin37(approx) sin(θ)cos(θ)-sin(37)cos(θ)=0 cos(θ)[sin(θ)-sin(37)]=0 either cos(θ)=0 i.e, (θ)=90 or sin(θ)-sin(37)=0 i.e (θ)=37 or 143 :)

  42. campbell_st
    • one year ago
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    does that make sense..?

  43. MTALHAHASSAN2
    • one year ago
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    so 37 and 143 be the answer

  44. MTALHAHASSAN2
    • one year ago
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    but i have to convert that in radians

  45. campbell_st
    • one year ago
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    so find the 1st quadrant angle for theta and then 2nd quadrant is \[\pi - \theta\]

  46. campbell_st
    • one year ago
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    this online calculator will do it for you http://web2.0calc.com/ just select radians

  47. campbell_st
    • one year ago
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    then input the ratio

  48. MTALHAHASSAN2
    • one year ago
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    how

  49. campbell_st
    • one year ago
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    select radian press 2nd... then asin which is arcsin \[asin = \sin^{-1}\] then enter 0.6 and = what do you get?

  50. campbell_st
    • one year ago
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    oops should select Rad.. bottom left corner

  51. MTALHAHASSAN2
    • one year ago
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    ok

  52. campbell_st
    • one year ago
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    then 2nd asin (0.6) =

  53. MTALHAHASSAN2
    • one year ago
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    0.605

  54. imqwerty
    • one year ago
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    the solutions are 90 = pi/2 rad 37=37pi/180 rad 143=143pi/180 rad do the calculations and get the answer in radians

  55. campbell_st
    • one year ago
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    ummm not quite... I got 0.6435011.... can you just check again you have to press 2nd to before closing the brackets then =

  56. MTALHAHASSAN2
    • one year ago
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    i goted 0.64

  57. campbell_st
    • one year ago
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    great so that is the 1st quadrant answer to the nearest hundredth now the 2nd quadrant is \[\pi - 0.64\]

  58. MTALHAHASSAN2
    • one year ago
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    2.50

  59. MTALHAHASSAN2
    • one year ago
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    right

  60. MTALHAHASSAN2
    • one year ago
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    @campbell_st

  61. campbell_st
    • one year ago
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    yes.. that's it... well done

  62. MTALHAHASSAN2
    • one year ago
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    wait but at the back they have four solution

  63. MTALHAHASSAN2
    • one year ago
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    these two are the same but we are missing two

  64. campbell_st
    • one year ago
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    ok... so they are the values for sin now the values of cos you have \[\cos(\theta) = 0\] what angles make it true..?

  65. MTALHAHASSAN2
    • one year ago
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    90

  66. campbell_st
    • one year ago
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    here is the curve |dw:1434608602050:dw|

  67. campbell_st
    • one year ago
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    remember its in radians

  68. MTALHAHASSAN2
    • one year ago
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    pie/2

  69. campbell_st
    • one year ago
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    so where does the cos curve cut the horizontal axis..?

  70. campbell_st
    • one year ago
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    great, that's 1 and the other is..?

  71. MTALHAHASSAN2
    • one year ago
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    270 which is 3pie/2

  72. campbell_st
    • one year ago
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    that's correct... so you have 4 angles...

  73. MTALHAHASSAN2
    • one year ago
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    yes thanks a lot

  74. campbell_st
    • one year ago
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    glad to help

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