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anonymous

  • one year ago

I have solved a trig problem and want to see if I got the correct answer.

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    @jdoe0001

  3. jdoe0001
    • one year ago
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    so..what did you get for 7?

  4. anonymous
    • one year ago
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    so each question has multiple solutions

  5. jdoe0001
    • one year ago
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    right

  6. anonymous
    • one year ago
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    For question #7, the solutions I got were\[\pi/4,2\pi/3 \]

  7. anonymous
    • one year ago
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    and\[5\pi/12,2\pi/3n\]

  8. jdoe0001
    • one year ago
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    well. you may not need the 3n part, since you're only constrained to \((0,2\pi)\)

  9. anonymous
    • one year ago
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    ok

  10. jdoe0001
    • one year ago
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    \(\bf 2cos(3\theta)+\sqrt{2}=0\implies 2cos(3\theta)=-\sqrt{2} \\ \quad \\ cos(3\theta)=-\cfrac{\sqrt{2}}{2}\implies 3\theta=cos^{-1}\left( -\cfrac{\sqrt{2}}{2} \right) \\ \quad \\ 3\theta= \begin{cases} \frac{3\pi }{4}\\ \frac{5\pi }{4} \end{cases}\qquad \theta= \begin{cases} \frac{\pi }{4}\\ \frac{5\pi }{12} \end{cases}\)

  11. anonymous
    • one year ago
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    so..... I just have two solutions for #7, which are pi/4 and 5pi/12

  12. jdoe0001
    • one year ago
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    yeap, due to the range being only \((0,2\pi)\)

  13. anonymous
    • one year ago
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    ok then how do I do #8

  14. jdoe0001
    • one year ago
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    cosine is negative only in the 3rd and 4th quadrants, thus

  15. jdoe0001
    • one year ago
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    one sec on 8

  16. anonymous
    • one year ago
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    wait one second, im going to show you how I solved it and you can tell me if its right

  17. jdoe0001
    • one year ago
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    k

  18. anonymous
    • one year ago
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    alright so for question #8, I only got 1 solution, 5pi/6

  19. anonymous
    • one year ago
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    uhh.. Its hing to take me a long time to type an equation

  20. anonymous
    • one year ago
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    let me just post a picture... give me one sec

  21. jdoe0001
    • one year ago
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    ok

  22. jdoe0001
    • one year ago
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    but \(\bf sin\left( \theta-\frac{\pi }{3} \right)+1=2\implies sin\left( \theta-\frac{\pi }{3} \right)=1 \\ \quad \\ \theta-\frac{\pi }{3} =sin^{-1}(1)\implies \theta-\cfrac{\pi }{3}=\cfrac{\pi }{2}\implies \theta=\cfrac{\pi }{2} +\cfrac{\pi }{3} \\ \quad \\ \theta=\cfrac{5\pi }{6}\)

  23. jdoe0001
    • one year ago
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    so.. that one is ok how about 9)?

  24. anonymous
    • one year ago
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  25. anonymous
    • one year ago
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    sorry I took so long, I was having trouble typing in the equation

  26. anonymous
    • one year ago
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    im stuck on number 9 and 10

  27. jdoe0001
    • one year ago
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    tis ok... hmm ok... hold the mayo on 9

  28. anonymous
    • one year ago
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    what?

  29. jdoe0001
    • one year ago
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    hmm hold the mayo, just tomatoes and mustard, one sec :)

  30. anonymous
    • one year ago
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    ok...

  31. jdoe0001
    • one year ago
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    do you recall what is \(\bf 1^2 = ?\)

  32. anonymous
    • one year ago
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    yeah

  33. jdoe0001
    • one year ago
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    k

  34. anonymous
    • one year ago
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    1

  35. jdoe0001
    • one year ago
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    \(\bf 3cos(\theta)+3=2sin^2(\theta) \\ \quad \\ 3[cos(\theta)+1]=2sin^2(\theta) \\ \quad \\ 3[cos(\theta)+1]=2[{\color{brown}{ 1-cos^2(\theta)}}] \\ \quad \\ 3[cos(\theta)+1]=2[{\color{brown}{ 1^2-cos^2(\theta)}}] \\ \quad \\ 3[cos(\theta)+1]=2[{\color{brown}{(1-cos(\theta))(1+cos(\theta))}}]\) folow it so far?

  36. anonymous
    • one year ago
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    yeah

  37. jdoe0001
    • one year ago
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    notice, \(1=1^2\), thus we use the difference of squares

  38. anonymous
    • one year ago
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    alright makes sense

  39. jdoe0001
    • one year ago
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    ok, one sec

  40. jdoe0001
    • one year ago
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    \(\bf 3[cos(\theta)+1]=2[{(1-cos(\theta))(1+cos(\theta))}] \\ \quad \\ 3[cos(\theta)+1]=2[{(1-cos(\theta))(cos(\theta)+1)}] \\ \quad \\ 3[cos(\theta)+1]-2{(1-cos(\theta))(cos(\theta)+1)}=0 \\ \quad \\ {\color{brown}{ [cos(\theta)+1]}}[3-2(1-cos(\theta))]=0 \\ \quad \\ \begin{cases} cos(\theta)+1=0\implies cos(\theta)=-1\implies \theta=cos^{-1}(-1) \\ \quad \\ 3-2(1-cos(\theta))\implies 1-cos(\theta)=\cfrac{-3}{-2}\implies 1-\cfrac{3}{2}=cos(\theta) \end{cases}\) notice the red part, is the common factor

  41. jdoe0001
    • one year ago
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    keeping in mind that \(cos(\theta)+1 \iff 1+cos(\theta)\)

  42. jdoe0001
    • one year ago
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    thus \(\bf \begin{cases} \theta=cos^{-1}(-1)\\ 1-\cfrac{3}{2}=cos(\theta)\implies -\cfrac{1}{2}=cos(\theta)\implies cos^{-1}\left( -\cfrac{1}{2} \right)=\theta \end{cases}\)

  43. anonymous
    • one year ago
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    so the solutions are just theta

  44. jdoe0001
    • one year ago
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    cosine is -1 at \(\pi\) and -1/2 at the angles you found before, \(\cfrac{3\pi }{4},\cfrac{5\pi }{4}\)

  45. anonymous
    • one year ago
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    woah woah woah, from where did we get 3pi/4 and 5pi/4

  46. jdoe0001
    • one year ago
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    ehhe

  47. jdoe0001
    • one year ago
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    from the 2nd case, the one where cosine is -1/2

  48. anonymous
    • one year ago
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    so \[\cos^{-1} (\frac{ -1 }{ 2 })=\frac{ 3\pi }{ 4 },\frac{ 5\pi }{ 4 }\]

  49. anonymous
    • one year ago
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    @jdoe0001 so those are the two solutions

  50. jdoe0001
    • one year ago
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    yeap, plus the 1st case, where cosine was -1, which means \(\pi\)

  51. anonymous
    • one year ago
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    so there are 3 solutions

  52. anonymous
    • one year ago
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    pi,3pi/4,5pi/4

  53. jdoe0001
    • one year ago
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    yeap

  54. anonymous
    • one year ago
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    how about the last question

  55. jdoe0001
    • one year ago
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    one sec

  56. anonymous
    • one year ago
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    ok

  57. jdoe0001
    • one year ago
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    \(\bf cos(2\theta)+3=5cos(\theta) \\ \quad \\ {\color{brown}{ 2cos^2(\theta)-1 }}+3=5cos(\theta) \\ \quad \\ 2cos^2(\theta)-1+3-5cos(\theta)=0 \\ \quad \\ 2cos^2(\theta)-5cos(\theta)+2=0\impliedby \textit{notice, is a quadratic}\)

  58. jdoe0001
    • one year ago
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    was looking if it was factorable by integers, if not, then we'd do quadratic formula

  59. jdoe0001
    • one year ago
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    yeap, is factorable

  60. anonymous
    • one year ago
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    ok

  61. jdoe0001
    • one year ago
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    \(\bf 2cos^2(\theta)-5cos(\theta)+2=0\implies [2cos(\theta)-1][cos(\theta)-2] \\ \quad \\ \begin{cases} 2cos(\theta)-1\implies cos(\theta)=\cfrac{1}{2}\implies \theta=cos^{-1}\left( \cfrac{1}{2} \right) \\ \quad \\ cos(\theta)-2\implies cos(\theta)=2\implies \theta=cos^{-1}(2) \end{cases}\)

  62. jdoe0001
    • one year ago
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    and you can check the factoring with FOIL if you wish

  63. anonymous
    • one year ago
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    yes it also works with FOIL

  64. jdoe0001
    • one year ago
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    notice the 2nd case, it's angle whose cosine is 2 however cosine is always -1 or 1 or in between meanign the 2nd case can be tossed away, and only use the 1st one

  65. jdoe0001
    • one year ago
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    and of course, cosine is 1/2 at \(\bf \cfrac{\pi }{3},\cfrac{5\pi }{3}\)

  66. jdoe0001
    • one year ago
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    actually, had a missing 0 in the cases, but anyhow, they're = 0 =) \(\bf 2cos^2(\theta)-5cos(\theta)+2=0\implies [2cos(\theta)-1][cos(\theta)-2]=0 \\ \quad \\ \begin{cases} 2cos(\theta)-1=0\implies cos(\theta)=\cfrac{1}{2}\implies \theta=cos^{-1}\left( \cfrac{1}{2} \right) \\ \quad \\ cos(\theta)-2=0\implies cos(\theta)=2\implies \theta=cos^{-1}(2) \end{cases}\)

  67. jdoe0001
    • one year ago
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    anyway, have to dash :)

  68. anonymous
    • one year ago
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    WAIT SO WHAT ARE THE SOLUTIONS

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