Find the x-coordinates of all points on the curve $\displaystyle y = sin2x - 2sinx$ at which the tangent line is horizontal.

I don't know what to do and where to begin. :confused:

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- Oct 23rd 2007, 03:58 PMFalconPUNCH!X-Coordinates on Tangent Line
Find the x-coordinates of all points on the curve $\displaystyle y = sin2x - 2sinx$ at which the tangent line is horizontal.

I don't know what to do and where to begin. :confused: - Oct 23rd 2007, 04:08 PMJhevon
- Oct 23rd 2007, 04:10 PMFalconPUNCH!
- Oct 23rd 2007, 04:16 PMFalconPUNCH!
I got $\displaystyle y' = (cos(2)) (2x) - 2sin(x)cos(x) = 0 $ as the derivative and I pretty much don't know how to solve it for x.

- Oct 23rd 2007, 04:19 PMJhevon
- Oct 23rd 2007, 04:21 PMFalconPUNCH!
- Oct 23rd 2007, 04:32 PMJhevon
for the first part we use the chain rule:

$\displaystyle \frac d{dx}f(g(x)) = f'(g(x)) \cdot g'(x)$

for $\displaystyle \sin 2x$: $\displaystyle f(x) = \sin x$ and $\displaystyle g(x) = 2x$

the other piece is trivial. $\displaystyle \frac d{dx} \sin x = \cos x$ so $\displaystyle \frac d{dx} -2 \sin x = -2 \frac d{dx} \sin x = -2 \cos x$ - Oct 23rd 2007, 04:37 PMFalconPUNCH!
- Oct 23rd 2007, 04:39 PMJhevon
- Oct 23rd 2007, 04:46 PMFalconPUNCH!
- Oct 23rd 2007, 05:04 PMJhevon
- Oct 23rd 2007, 05:05 PMFalconPUNCH!
- Oct 23rd 2007, 05:07 PMJhevon
- Oct 23rd 2007, 05:14 PMFalconPUNCH!
- Oct 23rd 2007, 05:22 PMJhevon
are you taking me for a ride? or are you really trying? *suspicious smiley (we still need one of these)*

$\displaystyle 2 \cos^2 x - \cos x - 1 = 0$ ....this is a quadratic, see if you can factorize

$\displaystyle \Rightarrow (2 \cos x + 1)(\cos x - 1) = 0$

$\displaystyle \Rightarrow 2 \cos x + 1 = 0 \mbox{ or } \cos x - 1 = 0$

$\displaystyle \Rightarrow \cos x = - \frac 12 \mbox{ or } \cos x = 1$

what angles does this happen for?