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anonymous

  • one year ago

What values for theta(0≤theta≤2pi) satisfy the equation?

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

  3. anonymous
    • one year ago
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    Why don't you factor cos theta? :)

  4. anonymous
    • one year ago
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    If i factor cos theta i would get costheta(2sintheta+1)

  5. anonymous
    • one year ago
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    then cos theta=0 2sintheta+1=0

  6. anonymous
    • one year ago
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    Yes, then you would have two solutions, right?

  7. anonymous
    • one year ago
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    ye

  8. anonymous
    • one year ago
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    Oh, yes, you're right.

  9. anonymous
    • one year ago
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    I'm stuck at this part ;-;

  10. anonymous
    • one year ago
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    my textbook says use the unit circle but how??!

  11. anonymous
    • one year ago
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    You can still work on 2sintheta+1=0.

  12. anonymous
    • one year ago
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    it becomes sintheta=-1/2

  13. anonymous
    • one year ago
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    x.x

  14. anonymous
    • one year ago
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    Right so take a look at a unit circle.

  15. anonymous
    • one year ago
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    |dw:1438490613452:dw|

  16. anonymous
    • one year ago
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    :O

  17. anonymous
    • one year ago
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    First, find where cos theta=0.

  18. anonymous
    • one year ago
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    so costheta=0 is the line in the middle?

  19. anonymous
    • one year ago
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    Yes. They are pi/2 and 3pi/2.

  20. anonymous
    • one year ago
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    Then, find where sin theta =-1/2.

  21. anonymous
    • one year ago
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    hmm

  22. anonymous
    • one year ago
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    |dw:1438490799061:dw|

  23. anonymous
    • one year ago
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    No. You should look at the y-coordinate in finding for the sine theta,

  24. anonymous
    • one year ago
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    ohh i forgot (cos,sin)

  25. anonymous
    • one year ago
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    lol yes :D

  26. anonymous
    • one year ago
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    okie so it would be |dw:1438490900498:dw|

  27. anonymous
    • one year ago
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    Yes.

  28. anonymous
    • one year ago
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    :O so the answer is b!

  29. anonymous
    • one year ago
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    Correct. :D

  30. anonymous
    • one year ago
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    :D

  31. anonymous
    • one year ago
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    i might need help later :P but fur now bai!

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