# Thread: X-Coordinates on Tangent Line

1. ## X-Coordinates on Tangent Line

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

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

2. Originally Posted by FalconPUNCH!
Find the x-coordinates of all points on the curve $y = sin2x - 2sinx$ at which the tangent line is horizontal.

I don't know what to do and where to begin.
the tangent line is horizontal when the derivative is zero. find the derivative, set it equal to zero and solve for x

3. Originally Posted by Jhevon
the tangent line is horizontal when the derivative is zero. find the derivative, set it equal to zero and solve for x
Thanks I'll try that.

4. I got $y' = (cos(2)) (2x) - 2sin(x)cos(x) = 0$ as the derivative and I pretty much don't know how to solve it for x.

5. Originally Posted by FalconPUNCH!
I got $y' = (cos(2)) (2x) - 2sin(x)cos(x) = 0$ as the derivative and I pretty much don't know how to solve it for x.
that's because your derivative is incorrect

$y' = 2 \cos 2x - 2 \cos x$

6. Originally Posted by Jhevon
that's because your derivative is incorrect

$y' = 2 \cos 2x - 2 \cos x$
could you please explain how you got that because every time I try I get a different answer.

7. Originally Posted by FalconPUNCH!
could you please explain how you got that because every time I try I get a different answer.
for the first part we use the chain rule:

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

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

the other piece is trivial. $\frac d{dx} \sin x = \cos x$ so $\frac d{dx} -2 \sin x = -2 \frac d{dx} \sin x = -2 \cos x$

8. Originally Posted by Jhevon
for the first part we use the chain rule:

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

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

the other piece is trivial. $\frac d{dx} \sin x = \cos x$ so $\frac d{dx} -2 \sin x = -2 \frac d{dx} \sin x = -2 \cos x$
Oh I know what I did wrong. I forgot to take the -2 like this $-2 \frac d{dx} sinx$

9. Originally Posted by FalconPUNCH!
Oh I know what I did wrong. I forgot to take the -2 like this $-2 \frac d{dx} sinx$
ok. now set the derivative to zero and continue

10. Originally Posted by Jhevon
ok. now set the derivative to zero and continue
I got cos2x = cosx

11. Originally Posted by FalconPUNCH!
I got cos2x = cosx

$\cos 2x - \cos x = 0$

$\Rightarrow 2 \cos^2 x - 1 - \cos x = 0$

this is quadratic in cos(x). you can take it from here right?

12. Originally Posted by Jhevon
ok. now set the derivative to zero and continue
Originally Posted by Jhevon

$\cos 2x - \cos x = 0$

$\Rightarrow 2 \cos^2 x - 1 - \cos x = 0$

this is quadratic in cos(x). you can take it from here right?
To tell you the truth I can't remember any of the identities since I haven't used them in a year. I'll try and take it from there.

13. Originally Posted by FalconPUNCH!
To tell you the truth I can't remember any of the identities since I haven't used them in a year. I'll try and take it from there.
you should look them up. for double angles you must know that:

$\cos 2x = \cos^2 x - \sin^2 x = 2 \cos^2 x - 1 = 1 - 2 \sin^2 x$

$\sin 2x = 2 \sin x \cos x$

ok, continue. tell me if you have any problems

14. Originally Posted by Jhevon
you should look them up. for double angles you must know that:

$\cos 2x = \cos^2 x - \sin^2 x = 2 \cos^2 x - 1 = 1 - 2 \sin^2 x$

$\sin 2x = 2 \sin x \cos x$

ok, continue. tell me if you have any problems
Yeah I don't think I can do it. I'm just going in circles. I was never good with functions

15. Originally Posted by FalconPUNCH!
Yeah I don't think I can do it. I'm just going in circles. I was never good with functions
are you taking me for a ride? or are you really trying? *suspicious smiley (we still need one of these)*

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

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

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

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

what angles does this happen for?

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