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PRECALC

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solve for x in the interval zero to two pi.

2 sin 4x = 1


i got 5 pi/24 and pi/ 24 but my teacher said that there should be eight solutions. what would those be and how would i find it"

  • PRECALC -

    2 sin 4 x = 1 Divide both sides by 2

    sin 4 x = 1 / 2

    Take the inverse sine of both sides :

    4 x = sin ^ - 1 ( 1 / 2 )

    ________________________________________

    Remark :

    sin ^ - 1 ( 1 / 2 ) =

    pi / 6 + 2 pi n

    OR

    5 pi / 6 + 2 pi n


    Where n is an integer

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    4 x = pi / 6 + 2 pi n

    OR

    4 x = 5 pi / 6 + 2 pi n


    Divide both sides by 4

    x = pi / 24 + pi n / 2

    OR

    x = 5 pi / 24 + pi n / 2


    For n = 0

    x = pi / 24 + pi * 0 / 2

    x = pi / 24

    OR

    x = 5 pi / 24 + pi * 0 / 2

    x = 5 pi / 24


    For n = 1

    x = pi / 24 + pi * 1 / 2

    x = pi / 24 + pi / 2

    x = pi / 24 + 12 pi / 24

    x = 13 pi / 24

    OR

    x = 5 pi / 24 + pi * 1 / 2

    x = 5 pi / 24 + pi / 2

    x = 5 pi / 24 + 12 pi / 24

    x = 17 pi / 24


    For n = 2

    x = pi / 24 + pi * 2 / 2

    x = pi / 24 + pi

    x = pi / 24 + 24 pi / 24

    x = 25 pi / 24

    OR

    x = 5 pi / 24 + pi * 2 / 2

    x = 5 pi / 24 + pi

    x = 5 pi / 24 + 24 pi / 24

    x = 29 pi / 24


    For n = 3

    x = pi / 24 + pi * 3 / 2

    x = pi / 24 + 3 pi / 2

    x = pi / 24 + 3 * 12 pi / 24

    x = pi / 24 + 36 pi / 24

    x = 37 pi / 24

    OR

    x = 5 pi / 24 + pi * 3 / 2

    x = 5 pi / 24 + 3 pi / 2

    x = 5 pi / 24 + 3 * 12 pi / 24

    x = 5 pi / 24 + 36 pi / 24

    x = 41 pi / 24


    Solutions :

    x1 = pi / 24

    x2 = 5 pi / 24

    x3 = 13 pi / 24

    x4 = 17 pi / 24

    x5 = 25 pi / 24

    x6 = 29 pi / 24

    x7 = 37 pi / 24

    x8 = 41 pi / 24

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