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anonymous

  • one year ago

Solve each equation on the interval [0,2π): a. 4sin2x – 3 = 0 b. cos(3x) = -1

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  1. anonymous
    • one year ago
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    @freckles @IrishBoy123 @ganeshie8

  2. anonymous
    • one year ago
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    @peachpi

  3. anonymous
    • one year ago
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    @Jennjuniper

  4. anonymous
    • one year ago
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    @johnweldon1993

  5. anonymous
    • one year ago
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    idk how to do this hun. sorry.

  6. anonymous
    • one year ago
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    @e.mccormick

  7. anonymous
    • one year ago
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    it's okay

  8. anonymous
    • one year ago
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    @e.mccormick is offline btw

  9. anonymous
    • one year ago
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    @peachpi I'm assuming you don't know how to do this either??

  10. anonymous
    • one year ago
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    @TheSmartOne @whpalmer4

  11. IrishBoy123
    • one year ago
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    a. \(4sin2x – 3 = 0\) thus: \(4sin2x = 3\) you can finish that off with your calculator. solve for '2x' first.

  12. anonymous
    • one year ago
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    Wait, so do I solve 4sin(2x) with my calculator? What about the " = 3" that you put. Should the answer to 4sin(2x) be my answer to "a"? What about b? @IrishBoy123

  13. anonymous
    • one year ago
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    @poopsiedoodle

  14. anonymous
    • one year ago
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    @sammixboo

  15. anonymous
    • one year ago
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    @GenTorr

  16. anonymous
    • one year ago
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    can yo u help with my questinon

  17. GenTorr
    • one year ago
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    i can try

  18. anonymous
    • one year ago
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    @logan6767 depends on what your question is...

  19. anonymous
    • one year ago
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    Which of the following points lie in the solution set to the following system of inequalities? y less than or greater to x - 5 y less than or greater to -x - 4

  20. anonymous
    • one year ago
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    (1, 10) (-1, 10) (10, 1) (1, -10)

  21. anonymous
    • one year ago
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    @dumbcow

  22. anonymous
    • one year ago
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    Sorry, can't solve it @logan6767

  23. anonymous
    • one year ago
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    @jim_thompson5910

  24. anonymous
    • one year ago
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    4 sin (2x) - 3 = 0 4 sin (2x) = 3 sin (2x) = 3/4 2x = arcsin (3/4) → this is the 1st quadrant solution for 2x x = ½ arcsin (3/4) 2x = π - arcsin (3/4) → this is the 2nd quadrant solution for 2x x = ½(π - arcsin (3/4)) If 0<x<2π, then 0<2x<4π, so other solutions for 2x are 2x = 2π + arcsin (3/4) x = π + ½ arcsin (3/4) 2x = 3π - arcsin (3/4) x = ½(3π - arcsin (3/4))

  25. anonymous
    • one year ago
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    Is this correct or are you guessing??

  26. anonymous
    • one year ago
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    and I'm out

  27. anonymous
    • one year ago
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    what about b?

  28. anonymous
    • one year ago
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    So a has 6 solutions?: 2x = 2π + arcsin (3/4) x = π + ½ arcsin (3/4) 2x = 3π - arcsin (3/4) x = ½(3π - arcsin (3/4)) x = ½ arcsin (3/4) x = ½(π - arcsin (3/4)) Or 4 solutions?: x = π + ½ arcsin (3/4) x = ½(3π - arcsin (3/4)) x = ½ arcsin (3/4) x = ½(π - arcsin (3/4)) @peachpi

  29. anonymous
    • one year ago
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    4

  30. anonymous
    • one year ago
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    okay, but what about b??

  31. anonymous
    • one year ago
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    The angle here is 3x, so the interval is [0, 6π). Pull out your unit circle and find the angle that has a cosine of -1. Then add 2π twice to get the other 2 solutions

  32. anonymous
    • one year ago
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    Do I add 2pi to the -1 or to the angle that has a cosine of -1?

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