# finding horizontal tangents using d/dx of trigs

• Oct 24th 2012, 08:12 AM
kingsolomonsgrave
finding horizontal tangents using d/dx of trigs
Attachment 25388

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?
• Oct 24th 2012, 09:29 AM
Siron
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 $\displaystyle k \in \mathbb{Z}$