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anonymous

  • one year ago

Find all solutions in the interval [0, 2π). (sin x)(cos x) = 0

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  1. anonymous
    • one year ago
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    Make sin x = 0 and cos x =0.

  2. anonymous
    • one year ago
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    @mathway can u help me with a prob

  3. anonymous
    • one year ago
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    okay, then what do i do?

  4. anonymous
    • one year ago
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    Look at a unit circle and find radians/degrees that has cos x=0 or sin x=0,

  5. anonymous
    • one year ago
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    Remember that in (x,y), the x is the cosine and y is the sine.

  6. anonymous
    • one year ago
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    which quadrant do i look in?

  7. anonymous
    • one year ago
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    All because the interval is [0,2pi).

  8. LynFran
    • one year ago
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    |dw:1438148609046:dw|

  9. LynFran
    • one year ago
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    |dw:1438148715430:dw|

  10. anonymous
    • one year ago
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    okay, would cos x = 0 x= pi/2, 3pi/2 ?

  11. anonymous
    • one year ago
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    Yes. :)

  12. LynFran
    • one year ago
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    correct the 2npi is actually the period...

  13. anonymous
    • one year ago
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    Now you need to find sin x= 0.

  14. anonymous
    • one year ago
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    Are you taking a class on FLVS?

  15. anonymous
    • one year ago
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    would then would sin x = 0 x= 0, pi, 2pi ? and yes.

  16. anonymous
    • one year ago
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    Yes! This is in Module 5 in Pre-Calculus? :D

  17. anonymous
    • one year ago
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    yes it is lol

  18. anonymous
    • one year ago
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    Because I had this question and asked here 2 days ago!

  19. anonymous
    • one year ago
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    really? coincidence much lol

  20. anonymous
    • one year ago
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    Ikr, anyway, so what is the solution? x=????

  21. anonymous
    • one year ago
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    x = 0, pi/2, pi, 3pi/2 ?

  22. anonymous
    • one year ago
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    Great job!

  23. anonymous
    • one year ago
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    thank you so much for all of your help! much appreciated!!

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