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Math Help - Trigonometric Equations

  1. #1
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    Post Trigonometric Equations

    Can somebody help me with these?

    Solve the equation:
    1. tan3x (tanx - 1)=0

    Find all solutions of the equation in the interval [0, 2π)
    2. 2sin2= 2 + cosx
    3. 2sin2x + 3sinx + 1 = 0

    Thank you so much!!!!!!!!!!!!!!!!!!!!
    Isabel
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cuteisa89 View Post
    Can somebody help me with these?

    Solve the equation:
    1. tan3x (tanx - 1)=0
    tan3x (tanx - 1) = 0
    => tan3x = 0 or tanx - 1 = 0

    => 3x = arctan0
    => x = (1/3)arctan0
    => x = k*pi for k an integer

    => tanx = 1
    => x = arctan(1)
    => x = pi/4 + k*pi for k an integer



    Find all solutions of the equation in the interval [0, 2π)
    2. 2sin2= 2 + cosx
    3. 2sin2x + 3sinx + 1 = 0
    are these 2's supposed to be squares?
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    I'll assume the bold 2's mean squares, since i got no response
    Quote Originally Posted by cuteisa89 View Post
    Find all solutions of the equation in the interval [0, 2π)
    2. 2sin2= 2 + cosx
    2sin^2x = 2 + cosx
    => 2(1 - cos^2x) = 2 + cosx
    => 2 - 2cos^2x = 2 + cosx
    => 2cos^2x + cosx = 0
    => cosx(2cosx + 1) = 0
    => cosx = 0 or 2cosx + 1 = 0

    cosx = 0
    => x = arcos0
    => x = pi/2 , 3pi/2 for x in [0, 2pi]

    2cosx + 1 = 0
    => x = arccos(-1/2)
    => x = 2pi/3 , 4pi/3 for x in [0, 2pi]

    3. 2sin2x + 3sinx + 1 = 0
    2sin^2x + 3sinx + 1 = 0

    this is just a quadratic equation cleverly disguised. You can see this if you replaced sinx with y for example, you would end up with 2y^2 + 3y + 1 = 0 so we will treat this as a quadratic and try to foil.

    => (2sinx + 1)(sinx + 1) = 0
    => 2sinx + 1 = 0 or sinx + 1 = 0

    2sinx + 1 = 0
    => x = arcsin(-1/2)
    => x = 7pi/6, 11pi/6 for x in [0, 2pi]

    sinx + 1 = 0
    => x = arcsin(-1)
    => x = 3pi/2 for x in [0, 2pi]
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  4. #4
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    Sorry I meant the 2 as a exponent but it didn't come out quite the right way. Thank you for all the help though... It is much appreciate it!

    Isabel
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cuteisa89 View Post
    Sorry I meant the 2 as a exponent but it didn't come out quite the right way. Thank you for all the help though... It is much appreciate it!

    Isabel
    well as you can see, i already solved it with 2 as an exponent

    it's fine as long as you understand it
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  6. #6
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    Quote Originally Posted by Jhevon View Post
    tan3x (tanx - 1) = 0
    => tan3x = 0 or tanx - 1 = 0

    => 3x = arctan0
    => x = (1/3)arctan0
    => x = k*pi for k an integer
    this last line should be:

    => x = (k/3) pi for k an integer

    => tanx = 1
    => x = arctan(1)
    => x = pi/4 + k*pi for k an integer


    are these 2's supposed to be squares?
    RonL
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    this last line should be:

    => x = (k/3) pi for k an integer



    RonL
    right, i forgot about the 1/3
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