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    solve the equation sin3xcosx=cos3xsinx for -180<x<180?

    Any ideas on how to go about this question?

    Thanks in advance!
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    (Original post by Lavanyaa)
    solve the equation sin3xcosx=cos3xsinx for -180<x<180?

    Any ideas on how to go about this question?

    Thanks in advance!
    Recall the double angle formula

    \displaystyle \sin(A-B) = \sin A\cos B - \cos A\sin B
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    Divide through by Cosx giving you Sin3x = cos3xTanx

    Now divide through by Cos3x giving Tan3x = Tanx

    Then use DeMoivres theorem to expand Tan3x

    Collect the terms and you will have an expression to factorise.

    EDIT: sin(A-B) expansion looks easier.
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    (Original post by uberteknik)
    Divide through by Cosx giving you Sin3x = cos3xTanx

    Now divide through by Cos3x giving Tan3x = Tanx

    Then use DeMoivres theorem to expand Tan3x

    Collect the terms and you will have an expression to factorise.
    That's the craziest advice I've come across today.

    Why go through all that when the double angle formula simplifies things immediately?

    Besides, we can't assume she's doing further maths.
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    (Original post by Indeterminate)
    That's the craziest advice I've come across today.

    Why go through all that when the double angle formula simplifies things immediately?

    Besides, we can't assume she's doing further maths.
    Hey, did you notice my edit?
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    (Original post by uberteknik)
    Hey, did you notice my edit?
    I think I was too quick to reply
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    (Original post by Indeterminate)
    Recall the double angle formula

    \displaystyle \sin(A-B) = \sin A\cos B - \cos A\sin B
    By using the double angle formula I got sin(2x)
    Is it possible to explain how you actually solve it?

    and Lol I don't take further maths
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    (Original post by Lavanyaa)
    By using the double angle formula I got sin(2x)
    Is it possible to explain how you actually solve it?

    and Lol I don't take further maths

    So you moved RHS to LHS of the original equation and noticed it was the identity of sin(A-B)

    What were you left with on the RHS of that move?

    So you now have sin2x= what?

    When you know that, then simply find the arcsin of both LHS and RHS

    Which gives you 2x = something

    The rest is just division.
 
 
 
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