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    1) Solve the following equation for t (theta), giving solutions in the interval 0 deg =< t =< 360 deg.
    a) sin 2t - root 3 cos 2t = 0.

    I don't know how to start this, because I can't see a use for any trig identity here Any ideas? :rolleyes:
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    By the time I go and grab a piece of paper and pen, someone will already have answered you
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    Try to move the -root3 cos 2t to the other side, and divide both sides by cost2t

    So you have tan2t=root3
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    Divide everything by cos2t ... should sort you out!
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    (Original post by cpdavis)
    Try to move the -root3 cos 2t to the other side, and divide both sides by cost2t

    So you have tan2t=root3
    But then you lose solutions :rolleyes:
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    (Original post by bcrazy)
    Divide everything by cos2t ... should sort you out!
    You then lose solutions =/
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    (Original post by Faith01)
    By the time I go and grab a piece of paper and pen, someone will already have answered you
    Haha almost
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    (Original post by adil12)
    Haha almost
    I agree with the 2 people who already commented above..you don't lose solutions.
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    you dont lose solutions as 0 is not a solution.
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    (Original post by Faith01)
    I agree with the 2 people who already commented above..you don't lose solutions.
    Ohhhh of course my bad! Thanks everyone
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    Another way to do this problem would be to write it in the form R\sin (2t-\alpha), but it seems like overkill for this question.

    (Original post by adil12)
    Ohhhh of course my bad! Thanks everyone
    FWIW in case you want to know why you don't lose solutions: when you divide through by \cos 2t you're not cancelling a common factor, so it doesn't "disappear" as such, since it gets "caught" by the \sin 2t.
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    (Original post by nuodai)
    Another way to do this problem would be to write it in the form R\sin (2t-\alpha), but it seems like overkill for this question.



    FWIW in case you want to know why you don't lose solutions: when you divide through by \cos 2t you're not cancelling a common factor, so it doesn't "disappear" as such, since it gets "caught" by the \sin 2t.
    Yeah it seems too complex for this question xD

    And I get what you mean, thank you for the additional info nuodai
 
 
 
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