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Math Help - Need help on a couple of Trig questions, assignment due tomorrow

  1. #1
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    Need help on a couple of Trig questions, assignment due tomorrow

    I have an assignment thats due tomorrow, heres a couple I'm not sure how to do.

    1)Factor & simplify: Tan^(3)x+8

    2) Sec^(2)x-2Tan^(2)x=0
    I got x= pi/4, 3pi/4, and 5pi/4 but I don't know if it's right

    3) Tanx=-sqrt(3)/3
    I got x=-pi/6 but not sure on answer

    If someone could please help me solve these I would really appreciate it. Thanks in advance.
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  2. #2
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    Quote Originally Posted by KURO View Post
    I have an assignment thats due tomorrow, heres a couple I'm not sure how to do.

    1)Factor & simplify: Tan^(3)x+8

    2) Sec^(2)x-2Tan^(2)x=0
    I got x= pi/4, 3pi/4, and 5pi/4 but I don't know if it's right

    3) Tanx=-sqrt(3)/3
    I got x=-pi/6 but not sure on answer

    If someone could please help me solve these I would really appreciate it. Thanks in advance.
    1)Factor & simplify: Tan^(3)x+8
    That could be a typo.
    I see that it should be tan^(3)x = 8
    Then, take the cube roots of both sides,
    tanX = 2 ......a positive tan value, so X must be in the 1st or 3rd quadrants.
    X = arctan(2)
    X = 1.10715 radians, in the 1st quadrant
    X = (pi +1.010715) = 4.24874 radians, in the 3rd quadrant
    So,
    X = 1.10715 and 4.24874 radians

    -----------------------------------
    2) Sec^(2)x-2Tan^(2)x = 0

    In the trig identity
    sin^2(X) +cos^2(X) = 1,
    Divide both sides by cos^2(X),
    tan^2(X) +1 = sec^2(X) -------**

    So,
    tan^2(X) +1 -2tan^2(X) = 0
    -tan^2(X) = -1
    tan^2(X) = 1
    tanX = +,-1

    When tanX = 1,
    positive tan value, so X is in the 1st and 3rd quadrants,
    X = arctan(1)
    X = pi/4 and (pi +pi/4) = 5pi/4
    or, X = pi/4, and 5pi/4

    When tanX = -1,
    negative tan value, so X is in the 2nd and 4th quadrants,
    X = arctan(-1)
    X = (pi -pi/4) = 3pi/4 in the 2nd quadrant
    X = (2pi -pi/4) = 7pi/4 in the 4th quadrant

    Therefore, X = pi/4, 3pi/4, 5pi/4 and 7pi/4

    ----------------------------------
    3) Tanx = -sqrt(3)/3

    If you divide both numerator and denominator by sqrt(3), that's the same as
    tanX = -1 / sqrt(3)
    Negative tan, so X is in the 2nd and 4th quadrants.
    X = arctan[-1 / sqrt(3)]
    X = (pi -pi/6) = 5pi/6 in the 2nd quadrant
    X = (2pi -pi/6) = 11pi/6 in the 4th quadrant
    So,
    X = 5pi/6 and 11pi/6
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  3. #3
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    Quote Originally Posted by ticbol View Post
    1)Factor & simplify: Tan^(3)x+8
    That could be a typo.
    I see that it should be tan^(3)x = 8
    Then, take the cube roots of both sides,
    tanX = 2 ......a positive tan value, so X must be in the 1st or 3rd quadrants.
    X = arctan(2)
    X = 1.10715 radians, in the 1st quadrant
    X = (pi +1.010715) = 4.24874 radians, in the 3rd quadrant
    So,
    X = 1.10715 and 4.24874 radians

    -----------------------------------
    2) Sec^(2)x-2Tan^(2)x = 0

    In the trig identity
    sin^2(X) +cos^2(X) = 1,
    Divide both sides by cos^2(X),
    tan^2(X) +1 = sec^2(X) -------**

    So,
    tan^2(X) +1 -2tan^2(X) = 0
    -tan^2(X) = -1
    tan^2(X) = 1
    tanX = +,-1

    When tanX = 1,
    positive tan value, so X is in the 1st and 3rd quadrants,
    X = arctan(1)
    X = pi/4 and (pi +pi/4) = 5pi/4
    or, X = pi/4, and 5pi/4

    When tanX = -1,
    negative tan value, so X is in the 2nd and 4th quadrants,
    X = arctan(-1)
    X = (pi -pi/4) = 3pi/4 in the 2nd quadrant
    X = (2pi -pi/4) = 7pi/4 in the 4th quadrant

    Therefore, X = pi/4, 3pi/4, 5pi/4 and 7pi/4

    ----------------------------------
    3) Tanx = -sqrt(3)/3

    If you divide both numerator and denominator by sqrt(3), that's the same as
    tanX = -1 / sqrt(3)
    Negative tan, so X is in the 2nd and 4th quadrants.
    X = arctan[-1 / sqrt(3)]
    X = (pi -pi/6) = 5pi/6 in the 2nd quadrant
    X = (2pi -pi/6) = 11pi/6 in the 4th quadrant
    So,
    X = 5pi/6 and 11pi/6
    Thanks for the help! But for Number 1, that isn't a typo its: Tan^(3)x+8. Any idea how to solve it?
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  4. #4
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    Quote Originally Posted by KURO View Post
    Thanks for the help! But for Number 1, that isn't a typo its: Tan^(3)x+8. Any idea how to solve it?
    If it were not a typo, then,
    tan^3(X) +8
    It is a sum of two cubes.
    = tan^3(X) +2^3
    = [tanX +2]*[tan^2(X) -2tanX +2^2]
    = (tanX +2)(tan^2(X) -2tanX +4)

    That's it. It's just factored.
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    Thanks!
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  6. #6
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    Quote Originally Posted by kuro View Post
    i have an assignment thats due tomorrow, heres a couple i'm not sure how to do.

    1)factor & simplify: Tan^(3)x+8

    hint ... A^3 + b^3 = (a + b)(a^2 - ab + b^2)

    2) sec^(2)x-2tan^(2)x=0
    i got x= pi/4, 3pi/4, and 5pi/4 but i don't know if it's right

    missing x = 7pi/4

    3) tanx=-sqrt(3)/3
    i got x=-pi/6 but not sure on answer

    what interval for solutions? 0 to 2pi? If so, missing x = 5pi/6 and 11pi/6

    if someone could please help me solve these i would really appreciate it. Thanks in advance.
    1
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