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Math Help - How do you solve for x when 2sin(2x-60)=1?

  1. #1
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    How do you solve for x when 2sin(2x-60)=1?

    How do you solve for x for this equation: 2sin(2x-60)=1?

    where -2π ≤ x ≤ 2π

    [π=pi]

    I am unsure of where to start for this one
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Jd09 View Post
    How do you solve for x for this question: 2sin(2x-60)=1?

    where -2π ≤ x ≤ 2π

    I am unsure of where to start for this one

    2 \sin (2x - 60) = 1 \implies \sin (2x - 60) = \frac 12 \implies 2x - 60 = \sin^{-1} \left( \frac 12 \right) \implies x = \frac {\sin^{-1}(1/2) + 60}2

    so now what? what are all the solutions? which have x in the given range?
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  3. #3
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    Is this correct?

    (sin-1)sin(2x-60)=1/2(sin-1)

    (2x-60)=30
    x=45

    ??
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Jd09 View Post
    Is this correct?

    (sin-1)sin(2x-60)=1/2(sin-1)

    (2x-60)=30
    x=45

    ??
    that is correct...is that the only solution? you want -360^o \le x \le 360^o remember?
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  5. #5
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    135...

    and would -135 work as well?
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  6. #6
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    How do you solve for x when 2sin(2x-60)=1?

    Thank you for your help!
    Last edited by mr fantastic; April 7th 2009 at 10:07 PM. Reason: Moved the rest of the post to a new thread
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