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  1. #1
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    Tri

    Solve for x , 0≤x≤360. Give exact values for x
    tan (3x-(pie/3))=-1
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  2. #2
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    Quote Originally Posted by rachael
    Solve for x , 0≤x≤360. Give exact values for x
    tan (3x-(pie/3))=-1
    The interval for the answer is in degrees, while the pi/3 is in radians. Umm.

    Here is one way.

    tan[3x -pi/3] = -1

    Use the trig identity
    tan(A-B) = (tanA -tanB)/(1 +tanA*tanB)

    So,
    tan[3x -pi/3] = -1
    [tan(3x) -tan(pi/3)]/[1 +tan(3x)*tan(pi/3)] = -1
    tan(3x) -tan(pi/3) = -1*[1 +tan(3x)*tan(pi/3)] ------------(i)

    tan(pi/3) = tan(60 degrees) = sqrt(3)/1 = sqrt(3), so, to continue from (i),

    tan(3x) -sqrt(3) = -1 -tan(3x)*sqrt(3)
    Isolate the tan(3x)-terms,
    tan(3x) +sqrt(3)*tan(3x) = -1 +sqrt(3)
    Factor out the tan(3x) in the LHS,
    tan(3x)*[1 +sqrt(3)] = -1 +sqrt(3)
    tan(3x) = [-1 +sqrt(3)]/[1 +sqrt(3)]
    Rearranging the RHS,
    tan(3x) = [sqrt(3) -1]/[sqrt(3) +1]
    Rationalize the denominator of the RHS,
    Multiply both numerator and denominator of the RHS by the conjugate of the denominator, which is [sqrt(3) -1],

    Here, we use (a-b)^2 = a^2 -2ab +b^2, and (a+b)(a-b) = a^2 -b^2,

    tan(3x) = [3 -2sqrt(3) +1]/[3 -1]
    tan(3x) = [4 -2sqrt(3)]/[2]
    tan(3x) = 2 -sqrt(3)
    tan(3x) = 0.267949192

    A positive tangent value. That means angle 3x is in the 1st or 3rd quadrant.

    3x = arctan(0.267949192)
    3x = 15 degrees

    That also means that
    3x = 15deg ----in the 1st quadrant, or,
    3x = 180+15 = 195deg ----in the 3rd quadrant.

    Therefore,
    x = 15/3 = 5 deg ----------------answer.
    or, x = 195/3 = 65 deg ----------answer.

    ----------------------------------------------------
    Check.

    When x = 5deg,
    tan(3x -pi/3) = -1
    tan(3*5deg -60deg) =? -1
    tan(15deg -60deg) =? -1
    tan(-45deg) =? -1
    -1 =? -1
    Yes, so, OK.

    When x = 65deg,
    tan(3x -pi/3) = -1
    tan(3*65deg -60deg) =? -1
    tan(195deg -60deg) =? -1
    tan(135deg) =? -1
    -1 =? -1
    Yes, so, OK

    -----------------------------
    Ooppss, I forgot that x is within the interval [0,360 degrees]

    Since what was found so far were 2 acute angles only---5 and 65 degrees---, there are more values for x in the given interval or bounds or domain of x.

    Since the "angle" in the question is 3x, and one revolution is 360 deg, then 360/3 = 120deg. So, keep on adding 120deg to the 5deg and 65deg for more x-values until the interval is exhausted.
    Thus,
    x = 5, 65, 125, 185, 245, and 305 degrees ----answers, all five of them.

    Check the x=305deg, at least,....
    When x = 305deg,
    tan(3x -pi/3) = -1
    tan(3*305deg -60deg) =? -1
    tan(915deg -60deg) =? -1
    tan(855deg) =? -1
    -1 =? -1
    Yes, so, OK
    Last edited by ticbol; February 20th 2006 at 05:19 AM. Reason: more angles for x within the interval [0,360]
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  3. #3
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    Quote Originally Posted by rachael
    Solve for x , 0≤x≤360. Give exact values for x
    tan (3x-(pie/3))=-1
    If \tan x=-1 then all solutions must be,
    x=-\frac{\pi}{4}+\pi k where k is some integer.

    Thus, \tan \left(3x-\frac{\pi}{3}\right)=-1
    becomes, 3x-\frac{\pi}{3}=-\frac{\pi}{4}+\pi k.
    Thus,
    x=\frac{\pi}{36}+\frac{\pi}{3}k
    But we need that,
    0\leq x\leq 2\pi, thus,
    0\leq \frac{\pi}{36}+\frac{\pi}{3}k\leq 2\pi
    Thus, by solving the inequality.
    -.083\bar 3=-\frac{1}{12}\leq k\leq\frac{71}{12}= 5.91\bar 6
    Thus, since k is integer we have that.
    k=0,1,2,3,4,5.
    Thus, since x=\frac{\pi}{36}+\frac{\pi}{3} k=-5^o+60^ok
    We have that,
    x=5^o
    x=65^o
    x=125^o
    x=185^o
    x=245^o
    x=305^o
    One for each value of k.
    Q.E.D.
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