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    from the equation tan22X=3/2 how do I find tanX
    Thanks
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    (Original post by daviem)
    from the equation tan22X=3/2 how do I find tanX
    Thanks
    You have to use the double angle identity for tan(2x)
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    (Original post by lizard54142)
    You have to use the double angle identity for tan(2x)
    sorry, I don't understand
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    (Original post by lizard54142)
    You have to use the double angle identity for tan(2x)
    could you explain further?
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    square root both sides

    use the identity for tan2x

    http://www.uhigh.ilstu.edu/math/sond...s%20pt%202.htm
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    (Original post by daviem)
    sorry, I don't understand
    The identity is:

    tan(A \pm B) = \dfrac{tanA \pm tanB}{1 \mp tanAtanB}

    In your case, A = B
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    (Original post by lizard54142)
    The identity is:

    tan(A \pm B) = \dfrac{tanA \pm tanB}{1 \mp tanAtanB}

    In your case, A = B
    ok, thanks
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    (Original post by daviem)
    ok, thanks
    Do you see how to start now? The first thing to do would be get rid of the tan^2(2x) by square rooting, this will make things much easier
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    In all honesty I'm trying to a trig equation question (question 7b, aqa, pure core 2, 2014) and I'm just perplexed as to how to get the values of X.
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    (Original post by the bear)
    square root both sides

    use the identity for tan2x

    http://www.uhigh.ilstu.edu/math/sond...s%20pt%202.htm
    Thanks for that link
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    (Original post by daviem)
    In all honesty I'm trying to a trig equation question (question 7b, aqa, pure core 2, 2014) and I'm just perplexed as to how to get the values of X.
    Is this a different question? If so, can you post a screenshot?
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    (Original post by lizard54142)
    Is this a different question? If so, can you post a screenshot?
    Yeah hang on, I'll try
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    Here, the answer and the question
    Attached Files
  1. File Type: docx SR Trig exam question.docx (234.6 KB, 57 views)
  2. File Type: docx SR Trig exam answer.docx (256.5 KB, 50 views)
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    square root then find the solutions using your preferred method such as the graph but remember to double your range as its tan2x. e.g if it was 0<x<360 make it 0<2x<720 then find values of x by halfing all the angles in the range. note if you divide by two first then find angles between 0 and 360 then you will lose solutions.
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    (Original post by Nels98)
    square root then find the solutions using your preferred method such as the graph but remember to double your range as its tan2x. e.g if it was 0<x<360 make it 0<2x<720 then find values of x by halfing all the angles in the range. note if you divide by two first then find angles between 0 and 360 then you will lose solutions.
    This is the correct method, for C2 anyway.

    But don't forget, x is going to be - root(3/2) as well as +
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    for 7a you need to use the identity sin^2x+cos^2x=1
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    (Original post by Bobjim12)
    This is the correct method, for C2 anyway.

    But don't forget, x is going to be - root(3/2) as well as +
    I'm only in L6 but i hear it gets a bit more complex next year
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    (Original post by Nels98)
    I'm only in L6 but i hear it gets a bit more complex next year
    Yeah, like a previous poster put for some double angle identity, saw similar things in AQA's formula book, probably c3/c4 stuff.
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    (Original post by daviem)
    Here, the answer and the question
    Oh, well you don't need to find tan(x) as you originally thought in your original post! You just have to square root, and then use inverse tan. Notice though in part b) it is in terms of 2 \theta, but part 1 was just in terms of x
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    (Original post by Nels98)
    square root then find the solutions using your preferred method such as the graph but remember to double your range as its tan2x. e.g if it was 0<x<360 make it 0<2x<720 then find values of x by halfing all the angles in the range. note if you divide by two first then find angles between 0 and 360 then you will lose solutions.
    I did that, but I wasn't getting the right answers, I got my primary value as 0.886
 
 
 

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