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  1. #1
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    tri

    solve for x: sin (2x-(pie/6)) - cos (2x-(pie/6)) = 0
    0 ≤ x ≤ 2pie
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  2. #2
    MHF Contributor
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    Quote Originally Posted by rachael
    solve for x: sin (2x-(pie/6)) - cos (2x-(pie/6)) = 0
    0 ≤ x ≤ 2pie
    Here is one way.

    sin(2x -pi/6) -cos(2x -pi/6) = 0
    sin(2x -pi/6) = cos(2x -pi/6)
    Divide both sides by cos(2x -pi/6),
    tan(2x -pi/6) = 1

    A positive tangent value. Meaning, angle (2x -pi/6) is in the 1st or 3rd quadrants.

    (2x -pi/6) = arctan(1)
    (2x -pi/6) = pi/4 ---------------in the 1st quadrant,
    or,
    (2x -pi/6) = pi +pi/4 = 5pi/4 ----in the 3rd quadrant.

    So,
    2x -pi/6 = pi/4
    2x = pi/4 +pi/6
    2x = (6pi +4pi)/(4*6)
    2x = 10pi/24
    x = 10pi/(24*2) = 5pi/24 = (5/24)pi radians, or 37.5 degrees -----answer.

    Or,
    2x -pi/6 = 5pi/4
    2x = 5pi/4 +pi/6
    2x = (6*5pi +4pi)/(4*6)
    2x = 34pi/24
    x = 34pi/(24*2) = 17pi/24 = (17/24)pi rad, or 127.5 deg -----answer.

    -----------------------------
    Check, in degrees, for easier use of calculator,

    When x = 37.5 degrees,
    sin(2x -pi/6) -cos(2x -pi/6) = 0
    sin(2*37.5deg -30deg) -cos(2*37.5deg -30deg) =? 0
    sin(75deg -30deg) -cos(75deg -30deg) =? 0
    sin(45deg) -cos(45deg) =? 0
    1/sqrt(2) -1/sqrt(2) =? 0
    0 =? 0
    Yes, so, OK.

    When x = 127.5 degrees,
    sin(2x -pi/6) -cos(2x -pi/6) = 0
    sin(2*127.5deg -30deg) -cos(2*17.5deg -30deg) =? 0
    sin(255deg -30deg) -cos(255deg -30deg) =? 0
    sin(225deg) -cos(225deg) =? 0
    -0.7071 -(-0.7071) =? 0
    0 =? 0
    Yes, so, OK.
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  3. #3
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    Quote Originally Posted by rachael
    solve for x: sin (2x-(pie/6)) - cos (2x-(pie/6)) = 0
    0 ≤ x ≤ 2pie
    The similar idea here is used as I posted in your other thread.
    If \tan x=1 then all solutions have form,
    x=\frac{\pi}{4}+\pi k

    But first do what ticbol says to do divide by cosine.
    Thus,
    \tan\left( 2x-\frac{\pi}{6}\right)=1
    Thus, all solutions have form,
    2x-\frac{\pi}{6}=\frac{\pi}{4}+\pi k
    Thus,
    x=\frac{5\pi}{24}+\frac{\pi}{2}=12.5^o+90^ok
    But, 0^o\leq x\leq 360^o
    Thus,
    0^o\leq12.5^o+90^ok\leq 360
    Thus, (solving the inequality)
    -.13\bar 8\leq k\leq 3.86\bar 1.
    Thus,
    k=0,1,2,3 because k is an integer.
    Thus,
    x=12.5^o
    x=102.5^o
    x=192.5^o
    x=282.5^o
    Each one for each value of k.
    Q.E.D.
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  4. #4
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    thank you
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