finding horizontal tangents using d/dx of trigs

**kingsolomonsgrave** · October 24th 2012, 08:12 AM

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?

**Siron** · October 24th 2012, 09:29 AM

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

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

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

where $k \in \mathbb{Z}$

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