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Thread: Solving sin(pi/3-2x) = sin(x-pi/6)

  1. March 13th 2010, 12:28 AM #1
    thex10
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    Unhappy Solving sin(pi/3-2x) = sin(x-pi/6)

    Hi!

    Having problems getting correct answers for the equation
    sin(pi/3-2x) = sin(x-pi/6)

    Could someone help and point me to the right direction with this one ?

    Thanks!
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  2. March 13th 2010, 12:58 AM #2
    Grandad
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    Hello thex10

    Welcome to Math Help Forum!
    Quote Originally Posted by thex10 View Post
    Hi!

    Having problems getting correct answers for the equation
    sin(pi/3-2x) = sin(x-pi/6)

    Could someone help and point me to the right direction with this one ?

    Thanks!
    The general solution of the equation:
    \sin A = \sin B
    is:
    A = n\pi + (-1)^nB, n = 0 , \pm 1, \pm 2, ...
    So for instance,
    When n = 0,\; A = B

    When n = 1,\; A = \pi - B

    When n = 2,\; A = 2\pi + B

    ... and so on
    So to solve the equation
    \sin(\pi/3-2x) = \sin(x-\pi/6)
    you first write down the general solution, as above, where
    A = \pi/3 - 2x
    and
    B = x - \pi/6
    which is:
    \pi/3 - 2x =n\pi +(-1)^n(x - \pi/6)
    and then see what happens when you take particular values for n.

    Can you continue from here?

    Grandad
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  3. March 13th 2010, 01:24 AM #3
    thex10
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    Got it!

    Thank you very much!
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