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STEP I 2003 q

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    I was looking through a solution someone on here posted to a q and I'm not sure I follow...


    Can someone pls explain the highlighted step and how you'd go on from there?

    Thank you!

    (Original post by Horizontal 8)
    STEP I question 3

     2cos\left(\frac{x}{2}\right) = cosx \iff 2cos\left(\frac{x}{2}\right) = 2cos^2\left(\frac{x}{2}\right)-1
     \iff 2cos^2\left(\frac{x}{2}\right)-2cos\left(\frac{x}{2}\right)-1 =0

     \displaystyle \iff cos\left(\frac{x}{2}\right) = \frac{2 \pm  \sqrt{12}}{4} = \frac{1 \pm  \sqrt{3}}{2}

    But  \displaystyle \left|\frac{1 + \sqrt{3}}{2} \right|>1
    Hence no solutions arise from this.

    Also,  \displaystyle \frac{1 - \sqrt{3}}{2} < 0

    So we use  \displaystyle \frac{\sqrt{3}-1}{2} to find the other angles.

    Cosx is -ve for  \displaystyle \frac{\pi}{2}&lt;x&lt;\frac{3 \pi}{2} when considering 0<x<2pi

    Therefore the angles that we want are:

     \displaystyle \frac{x}{2} = (2n+1)\pi \pm cos^{-1}\left(\frac{\sqrt{3} -1}{2} \right)

     \implies \displaystyle x = (4n+2)\pi \pm 2cos^{-1}\left(\frac{\sqrt{3} -1}{2} \right)

    And this is equal to  \boxed{\displaystyle (4n+2)\pi \pm 2 \phi \right)}

    As required.
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    Notice that \cos\theta \equiv \cos(-\theta)
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    hmm I still don't understand why you wouldn't just use the cosx instead of cos(-x) to work through the remainder of the question???
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    (Original post by shanban)
    hmm I still don't understand why you wouldn't just use the cosx instead of cos(-x) to work through the remainder of the question???
    I think it is because the interval you are dealing with in the equation is [0;\pi/2]. Since \displaystyle \frac{1 - \sqrt{3}}{2} &lt; 0, you rule it out and use the equivalent positive value, which is  \displaystyle \frac{\sqrt{3}-1}{2}
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    (Original post by z0tx)
    I think it is because the interval you are dealing with in the equation is [0;\pi/2]. The absolute value gives two possibilities. Since \displaystyle \frac{1 - \sqrt{3}}{2} &lt; 0, you rule it out and use the other (positive) value, which is  \displaystyle \frac{\sqrt{3}-1}{2}
    oh of course! got it, thank you

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