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    aussie_kris's Avatar
    aussie_kris Posts: 6, Reputation: 1
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    #1

    May 8, 2005, 11:58 PM
    Trig question
    Hey

    how do I solve

    2cos^2($) + sin ($) = 1

    Any help would be great
    aussie_kris's Avatar
    aussie_kris Posts: 6, Reputation: 1
    New Member
     
    #2

    May 9, 2005, 03:15 AM
    Hey I need to solve it for theta -pie to pie
    aussie_kris's Avatar
    aussie_kris Posts: 6, Reputation: 1
    New Member
     
    #3

    May 9, 2005, 03:57 AM
    Hey thanks

    I think that's what I need

    also any hints with expressing cos3($) in terms of sin and sin3($) in terms of cos??
    aussie_kris's Avatar
    aussie_kris Posts: 6, Reputation: 1
    New Member
     
    #4

    May 9, 2005, 04:02 AM
    How did u get 5.176?? I got -63.435
    shanus's Avatar
    shanus Posts: 17, Reputation: 0
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    #5

    May 9, 2005, 07:47 AM
    ooops!
    Ahhhh... I hate to burst eveyone's bubble here, but...

    Cos2($) is not equal to 2x/r

    i.e.. Cos2($) is not equal to 2Cos($)

    I think a lot of Mathematicians would be upset if this were true ;)
    MathMaven53's Avatar
    MathMaven53 Posts: 20, Reputation: 2
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    #6

    Jun 29, 2005, 01:38 PM
    Trig question
    Let's express the equation in terms of Sin(x)

    We have 2 Cos^2 (x) + Sin(x) =1

    We use Cos^2 (x) = 1 - Sin^2(x)

    After substituting and simplifying we get the equation

    2 Sin^2(x) - Sin(x) - 1 = 0

    Factor this into

    (2Sin(x) + 1)(Sin(x) -1) = 0

    Now set each factor equal to zero,solve for Sin(x) and then x

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