C3 trig addition formulae equation

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    I am trying to solve for theta but have ended up with this. What am I supposed to do now?


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    (Original post by jessyjellytot14)

    I am trying to solve for theta but have ended up with this. What am I supposed to do now?


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    Divide both sides by cosine to get \tan \theta
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    (Original post by jessyjellytot14)
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    I am trying to solve for theta but have ended up with this. What am I supposed to do now?


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    On line 2:
    \cos \theta \cos 60 = \sin \theta(1-\sin60)
    \frac{\cos 60}{1- \sin 60} = \tan \theta
    I think you can take it from here
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    (Original post by RDKGames)
    Divide both sides by cosine to get \tan \theta
    the way it was simplified at the end would give a  \frac{1}{\cos \theta} as well
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    (Original post by 123Master321)
    the way it was simplified at the end would give a  \frac{1}{\cos \theta} as well
    No it wouldn't. I assume OP notices the mistake of forgetting the brackets. Otherwise, that is something she would quickly realise.
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    (Original post by RDKGames)
    No it wouldn't. I assume OP notices the mistake of forgetting the brackets. Otherwise, that is something she would quickly realise.
    Oh yh my bad, I just assumed the working was right and it got simplified poorly
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    (Original post by RDKGames)
    Divide both sides by cosine to get \tan \theta
    (Original post by 123Master321)
    On line 2:
    \cos \theta \cos 60 = \sin \theta(1-\sin60)
    \frac{\cos 60}{1- \sin 60} = \tan \theta
    I think you can take it from here
    I tried both of these methods and they both got the same answers
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    (Original post by jessyjellytot14)
    I tried both of these methods and they both got the same answers
    That's because these aren't different methods.
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    (Original post by RDKGames)
    That's because these aren't different methods.
    I know but the first method doesn't involve factorising so they look slightly different when written down.
 
 
 
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