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Thread: finding horizontal tangents using d/dx of trigs

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    finding horizontal tangents using d/dx of trigs

    finding horizontal tangents using d/dx of trigs-screen-shot-2012-10-24-12.11.22-pm.png

    What is going on in the last two steps?

    Is it that 2x must be the value that makes sin theta = sqrt(3)/2 and so it must be pi/3 or 2pi/3 or even integer multiples thereof and the second step we divide by two to get x?
    Last edited by kingsolomonsgrave; Oct 24th 2012 at 08:52 AM.
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: finding horizontal tangents using d/dx of trigs

    Yes, that's what happens.
    $\displaystyle \sin 2x = \frac{\sqrt{3}}{2}}$

    Let $\displaystyle 2x=u$ then we obtain the equation $\displaystyle \sin(u) = \frac{\sqrt{3}}{2} \Leftrightarrow \sin u = \sin\left(\frac{\pi}{3}\right)$
    We have two different solutions now:
    $\displaystyle u = \frac{\pi}{3}+2k\pi \Rightarrow 2x = \frac{\pi}{3}+2k\pi$
    $\displaystyle u = \pi - \frac{\pi}{3}+2k\pi \Rightarrow 2x = \frac{2\pi}{3}+2k\pi$

    which gives us (if we divide both sides by 2)
    $\displaystyle x = \frac{\pi}{6}+k\pi$
    $\displaystyle x= \frac{\pi}{3}+k \pi$

    where $\displaystyle k \in \mathbb{Z}$
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