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Math Help - Just one More :)

  1. #1
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    Post Just one More :)

    Ok..maybe not one more but three

    I'm supposed to find all solution of the equation in the inverval [0, 2pi).

    1. sin2x = - sqrt(3)/2.

    I tried solving this one. I have something like this: 2pi/3 + npi and 5pi/12 + npi.

    2. tan3x = 1

    3. Sec 4x = 2

    Thanks for all the help!

    Isabel
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  2. #2
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    Quote Originally Posted by cuteisa89 View Post
    Ok..maybe not one more but three

    I'm supposed to find all solution of the equation in the inverval [0, 2pi).

    1. sin2x = - sqrt(3)/2.
    It is negative, we are in the third and fourth quadrants.

    Thus, at angles 240 and 300.

    Thus,

    2x = 240 +180k or 300+180k

    Thus,

    x = 120 + 90k or 150+90k
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  3. #3
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    Hello, Isabel!

    Don't forget the interval . . .


    Find all solutions in the inverval [0, 2π)
    . . . . . . . . . . . . _
    1) .sin(2x) .= .-√3/2.
    . . . . . . .-. . .-. . . . . . . . . ._
    We know that: .sin(π/3) = √3/2
    . . and sine is negative in quadrants 3 and 4.

    Hence: .2x .= .4π/3, 5π/3, 10π/3, 11π/3

    Therefore: .x .= .2π/3, 5π/6, 5π/3, 11π/6



    2) .tan(3x) .= .1
    We know that: .tan(π/4) = 1
    . . and tangent is positive in quadrants 1 and 3.

    Hence: .3x .= .π/4, 5π/4, 9π/4, 13π/4, 17π/4, 21π/4

    Therefore: .x .= .π/12, 5π/12, 3π/4, 13π/12, 17π/12, 7π/4



    3) .sec(4x) .= .2

    We know that: .sec(π/3) = 2
    . . and secant is positive in quadrants 1 and 4.

    Hence: .4x .= .π/3, 5π/3, 7π/3, 11π/3, 13π/3, 17π/3, 19π/3, 23π/3

    Therefore: .x .= .π/12, 5π/12, 7π/12, 11π/12, 13π/12, 17π/12, 19π/12, 23π/12

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