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    Would someone be able to help me with some C3 Trig?

    I have 2 questions

    a) √3 sin2(x) + 2sin^2(x) = 1

    b) 4 tan2(x) tan(x) =1

    Thank you.
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    Have you used your knowledge of double angle identities?

    Edit: That probably isn't enough help. For the first one, express \sin^2 x in terms of \cos 2x.
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    I have but I'm still stuck with what to do with the 2sin^2 (x) - from a) and the tan(x)- from b)
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    (Original post by adelmoulton)
    I have but I'm still stuck with what to do with the 2sin^2 (x) - from a) and the tan(x)- from b)
    You have changed the wrong part in a) and you don't need to do anything the tan x in b) except multiply by it.
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    (Original post by adelmoulton)
    Would someone be able to help me with some C3 Trig?

    I have 2 questions

    a) √3 sin2(x) + 2sin^2(x) = 1

    b) 4 tan2(x) tan(x) =1

    Thank you.
    do you mean √3 sin(2x)?

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    (Original post by Goods)
    do you mean √3 sin(2x)?

    Yes sorry I did.
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    (Original post by Mr M)
    You have changed the wrong part in a) and you don't need to do anything the tan x in b) except multiply by it.
    In a) I changed the 2sin^2x to (1- cos2x) and got √3 2sinxcosx + (1-cos2x) = 1

    Should I now divide by √3 or should I try and simplify?

    Thanks
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    (Original post by adelmoulton)
    In a) I changed the 2sin^2x to (1- cos2x) and got √3 2sinxcosx + (1-cos2x) = 1

    Should I now divide by √3 or should I try and simplify?

    Thanks
    I have already told you not to replace the sin 2x.

    Change it back and subtract 1 from each side.

    Now think of a Core 2 identity ...
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    (Original post by Mr M)
    I have already told you not to replace the sin 2x.

    Change it back and subtract 1 from each side.

    Now think of a Core 2 identity ...
    I'm sorry I still don't get it, please can you explain in more detail.
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    (Original post by adelmoulton)
    I'm sorry I still don't get it, please can you explain in more detail.
    \sqrt 3 \sin 2x + 1 - \cos 2x = 1

    \sqrt 3 \sin 2x = \cos 2x

    \displaystyle \frac{\sin 2x}{\cos 2x}  = \frac{1}{\sqrt 3}

    etc
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    (Original post by Mr M)
    \sqrt 3 \sin 2x + 1 - \cos 2x = 1

    \sqrt 3 \sin 2x = \cos 2x

    \displaystyle \frac{\sin 2x}{\cos 2x}  = \frac{1}{\sqrt 3}

    etc
    Thank you - I appreciate the help
 
 
 
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