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    Solve for 0< θ<360

    Cos(θ-30) = 2Sinθ

    Im stuck please help, i tried expanding the brackets and solving but it seems wrong
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    (Original post by Rockyboy)
    Solve for 0< θ<360

    Cos(θ-30) = 2Sinθ

    Im stuck please help, i tried expanding the brackets and solving but it seems wrong
    How so?

    \cos(\theta - 30)=\cos(\theta)\cos(30)+\sin( \theta)\sin(30) Then you can get \tan(\theta) out of it.
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    Dividing by an expression with just sin and cos to get tan is a handy trick like knowing lne=1. I forgot it was best to divide by cos when I was doing a past paper and did the whole Rsin(x+a) thing when you didn't have to.
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    (Original post by RDKGames)
    How so?

    \cos(\theta - 30)=\cos(\theta)\cos(30)+\sin( \theta)\sin(30) Then you can get \tan(\theta) out of it.
    Yes but what to i do after. Cuz i got

    1+tan(θ)tan(30) =2sin( θ)
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    (Original post by Rockyboy)
    Yes but what to i do after. Cuz i got

    1+tan(θ)tan(30) =2sin( θ)
    That's wrong.

    \cos(\theta)\cos(30)+\sin( \theta)\sin(30) = 2\sin(\theta)

    So \cos(\theta)\cos(30) = \sin( \theta)[2-\sin(30)]

    Thus \tan(\theta)=\frac{\cos(30)}{2-\sin(30) }
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    (Original post by black1blade)
    Dividing by an expression with just sin and cos to get tan is a handy trick like knowing lne=1. I forgot it was best to divide by cos when I was doing a past paper and did the whole Rsin(x+a) thing when you didn't have to.
    dont you lose a solution if you divide by sin/cos or does that not apply to this question?
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    (Original post by RDKGames)
    That's wrong.

    \cos(\theta)\cos(30)+\sin( \theta)\sin(30) = 2\sin(\theta)

    So \cos(\theta)\cos(30) = \sin( \theta)[2-\sin(30)]

    Thus \tan(\theta)=\frac{\cos(30)}{2-\sin(30) }
    I got it thx
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    Never mind im being stupid. i get it now thank you very much
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    (Original post by ForestShadow)
    dont you lose a solution if you divide by sin/cos or does that not apply to this question?
    Say if you had sinx+cosx=0 then the quickest approach is to divide by cosx. However yeah if possible you should always fractorise so if it was cos^2(x)+sinxcosx=0 you would factorise first to get cosx(cosx+sinx)=0 then say cosx=0 and cosx+sinx=0.
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    (Original post by black1blade)
    Say if you had sinx+cosx=0 then the quickest approach is to divide by cosx. However yeah if possible you should always fractorise so if it was cos^2(x)+sinxcosx=0 you would factorise first to get cosx(cosx+sinx)=0 then say cosx=0 and cosx+sinx=0.
    ok thanks!
 
 
 
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