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    Holaa!

    I'm probably being dense again, but I've re-written this out a thousand times now and can't see what I'm supposed to do, so any help would be great, thanks!

    Solve the equation: (3cos2x)/(cos2x+1)=2(tanx+1) ; range 0-360

    I'm assuming you expand each cos2x, but then every time I try and rearrange what I have left, I end up with disgustingly long expressions and I'm sure it must be simpler than this!

    Thank you!
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    (Original post by Lainathiel)
    Holaa!

    I'm probably being dense again, but I've re-written this out a thousand times now and can't see what I'm supposed to do, so any help would be great, thanks!

    Solve the equation: (3cos2x)/(cos2x+1)=2(tanx+1) ; range 0-360

    I'm assuming you expand each cos2x, but then every time I try and rearrange what I have left, I end up with disgustingly long expressions and I'm sure it must be simpler than this!

    Thank you!
    Cos2x + 1 should trigger which double angle formulae for cos you should be using (the one that cancels out nicely!).

    And then thinking about what that leaves you on the denominator for the left hand side, think about which one you should use for the top cos2x.
    Spoiler:
    Show
    You should be aiming for a quadratic equation in terms of tanx.
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    (Original post by SeanFM)
    Cos2x + 1 should trigger which double angle formulae for cos you should be using (the one that cancels out nicely!).

    And then thinking about what that leaves you on the denominator for the left hand side, think about which one you should use for the top cos2x.
    Spoiler:
    Show
    You should be aiming for a quadratic equation in terms of tanx.
    That's what I did before and still got confused; I'm at (3-6sin^2x)/(2cos^2x) which can then be written as 3/(2cos^2x)-3tan^2x, but then you still have the sec in there, which I don't know how I'm supposed to get rid of it!

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    (Original post by Lainathiel)
    That's what I did before and still got confused; I'm at (3-6sin^2x)/(2cos^2x) which can then be written as 3/(2cos^2x)-3tan^2x, but then you still have the sec in there, which I don't know how I'm supposed to get rid of it!

    Posted from TSR Mobile
    Your substitution for the denominator is correct.

    I'm not sure if you can work from the substitution you've done for the numerator, but instead of cos2x = 1 - 2sin^2(x), I would suggest using cos2x = (cos^2(x) - sin^2(x)).
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    (Original post by SeanFM)
    Your substitution for the denominator is correct.

    I'm not sure if you can work from the substitution you've done for the numerator, but instead of cos2x = 1 - 2sin^2(x), I would suggest using cos2x = (cos^2(x) - sin^2(x)).
    So, (3cos^2x)/(2cos^2x)-(3/2*tanx)=4tanx-4 - but I rearranged that and the quadratic is unsolvable?!

    I'm sorry for being a pain - it's been a long week!
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    (Original post by Lainathiel)
    So, (3cos^2x)/(2cos^2x)-(3/2*tanx)=4tanx-4 - but I rearranged that and the quadratic is unsolvable?!

    I'm sorry for being a pain - it's been a long week!
    It's okay :borat:

    I think you've half-multiplied the equation by 2 (it should be what you've got on the left hand side = 2tanx + 2 rather than 4tanx - 4).
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    (Original post by SeanFM)
    It's okay :borat:

    I think you've half-multiplied the equation by 2 (it should be what you've got on the left hand side = 2tanx + 2 rather than 4tanx - 4).
    Damn it - the reason it was unsolvable was because I had a quadratic of 3x^2+4x+7, not +1. *facepalm* I told you I was being dense!! Okay, I FINALLY have it. :')

    Thank you so much!
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    Ok so step 1:
    3(cos^2(X)-sin^2(X))/cos^2(X)-sin^2(X)+1

    Step 2:
    You need to use sin^2(X)+cos^2(X)=1
    This can rearrange to sin^2(X)=1-cos^2(X)
    Sub this in on the bottem half of the fraction to get
    3(cos^2(X)-sin^2(X))/cos^2(X)-(1-cos^2(X))+1

    step 3:
    Simplify the bottem
    3(cos^2(X)-sin^2(X))/2cos^2(X)

    Step 4:
    Split it all up to make it easier
    3(cos^2(X)/2cos^2(X))(-sin^2(X))/2cos^2(X))
    And simplify
    3(1/2)(-tan^2(X)/2)
    -3/2tan^2(X)

    Step 5:
    We know it equals 2(tanx+1)
    So set them equal
    -3/2tan^2(X)=2(tanx+1)
    Expand the right side
    -3/2tan^2(X)=2tanx+2
    This makes a quadratic
    3/2tan^2(X)+2tanx+2 = 0

    Step 6:
    Let tanx = a
    So 3/2a^2+2a+2=0
    And solve it like a normal quadratic equation (you might need to use the formula)
    I will let you do this bit

    Step 7:
    When you know what a equals you know what tanx equals
    Just do X=tan^-1(a) (inverse tan of what ever a was)
    Then find all the corisponding values within you set range

    Hope that helps!
    Let me know if you are still stuck
 
 
 
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