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    Please please help with the following:

    Solve to 3 sig. figs.:


    4tan2x+3cotxsec2x=0
    for 0<theta<2π
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    (Original post by Primrose96)
    Please please help with the following:

    Solve to 3 sig. figs.:


    4tan2x+3cotxsec2x=0
    for 0<theta<2π
    Well, we don't post solutions for people here, I'm afraid.

    Clue, use trig identities to convert it all to the same trig function. That makes it easier
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    (Original post by Primrose96)
    Please please help with the following:

    Solve to 3 sig. figs.:


    4tan2x+3cotxsec2x=0
    for 0<theta<2π
    You may find this post gives you an idea

    http://www.thestudentroom.co.uk/show....php?t=2462805
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    I have got this far. Is this correct and how can I solve it from here?
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    (Original post by Primrose96)
    I have got this far. Is this correct and how can I solve it from here?
    you are on the right lines converting to tan. But you should have started by getting rid of the zero. What you have now is less helpful.

    Step one is to move the 3cot x sec2x to the RHS.

    then do your converting as before

    you'll get a quartic in tan which can be reduced to a quadratic in tan2
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    Think I have it figured out now. Thank you so much
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    (Original post by Primrose96)
    Think I have it figured out now. Thank you so much
    no probs
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    Could I please have some help on another question:

    Solve in radians in the range 0<x<2π

    3sin2x=4cos2x

    I attempted to change cos2x to (1-2(sin^2)x) but I ended up with 8(sin^2)x + 3sin2x = 4
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    (Original post by Primrose96)
    Could I please have some help on another question:

    Solve in radians in the range 0<x<2π

    3sin2x=4cos2x

    I attempted to change cos2x to (1-2(sin^2)x) but I ended up with 8(sin^2)x + 3sin2x = 4
    remember tan is just sin/cos. That simplifies life considerably
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    Just finished this question. Haha I was getting worried coz I got stuck half way through. Anyway I'll give you a hint:

    Turn it all into tan. Then there will be a formula that can be factorised into double brackets
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    (Original post by kesupile)
    Just finished this question. Haha I was getting worried coz I got stuck half way through. Anyway I'll give you a hint:

    Turn it all into tan. Then there will be a formula that can be factorised into double brackets
    No. It's simpler than that
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    (Original post by Plato's Trousers)
    No. It's simpler than that
    Haha dude I'm talking about the problem that OP posted

    And I agree with you. The second one just requires you to remember that tan(x) is sin(x)/cos(x)
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    (Original post by kesupile)
    Haha dude I'm talking about the problem that OP posted

    And I agree with you. The second one just requires you to remember that tan(x) is sin(x)/cos(x)
    Oh right. Sorry. Yes, the original problem is just a quadratic
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    Am I along the right lines?

    3sin2x = 4cos2x

    let 2x=y

    3siny=4cosy

    siny/cosy = 4/3

    tany = 4/3

    tan(2x) = 4/3
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    (Original post by Primrose96)
    Am I along the right lines?

    3sin2x = 4cos2x

    let 2x=y

    3siny=4cosy

    siny/cosy = 4/3

    tany = 4/3

    tan(2x) = 4/3
    Yes. If we want to be really pedantic, then you should let y = 2x, not let 2x = y, but you are entirely correct in your logic.
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    (Original post by Primrose96)
    Am I along the right lines?

    3sin2x = 4cos2x

    let 2x=y

    3siny=4cosy

    siny/cosy = 4/3

    tany = 4/3

    tan(2x) = 4/3
    spot on
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    Prove the identities:

    cos2A = (1-tan2A) / (1+tan2A)

    For this is it better to work with the LHS or RHS?
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    (Original post by Primrose96)
    Prove the identities:

    cos2A = (1-tan2A) / (1+tan2A)

    For this is it better to work with the LHS or RHS?
    I'd start with the RHS and write tan in terms of sin and cos
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    Solve in the range 0<x<360

    tan2x - tanx = 0

    So far I have done 2tanx / (1-(tan^2)x) = tanx

    2tanx = tanx (1 - (tan^2)x)

    2 = 1 - ((tan^2)x)

    (tan^2)x = -1

    Where am I going wrong as you can't have a square root of a minus number?
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    (Original post by Primrose96)
    Solve in the range 0<x<360

    tan2x - tanx = 0

    So far I have done 2tanx / (1-(tan^2)x) = tanx

    2tanx = tanx (1 - (tan^2)x)

    2 = 1 - ((tan^2)x)

    (tan^2)x = -1

    Where am I going wrong as you can't have a square root of a minus number?
    Ok, this is driving me mad.

    It's clearly wrong since the answer is 0 and 180, but I'm damned if I can see your error!
 
 
 
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