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so first do you know how to find dy/dtheta and dx/dtheta ?

ok cool
now we need to set both of those things equal to 0 and solve for theta

can you solve either one or both or none?

I was able to get \[\sin \theta = 0 \]
\[\theta = \pi, 2\]

2 pi

are there *

That was where I was having trouble, since I am really rusty with my trig.

well hey does the problem say anything about restrictions on theta?

like is theta between 0 and 2pi
or does it not say anything about that?

It doesn't. :\

ok so there is an infinite amount of solutions

that just means n can be ...,-5,-4,-3,-2,-1,0,1,2,3,4,5,... so on in either direction

Hmm, sounds like things could get complicated...

do you know the period of sine is 2pi?

Yes.

That was what I got for that part. :)

it sounds like you should be looking for dy/dx =0 (not dy/theta or dx/dtheta)

but of course we can go the other way around to include the negatives

anyways I'm probably going to be lazy in continue using t instead of theta

so now we have cos(2t)=0

solve this first where t is in [0,2pi)

That's fine. Use t. cos (2t) = 0 is what is give me issues.

ok do you know how to solve cos(u)=0 ?

Wouldn't you take arccos of both sides?

Oh, so just keep adding or subtracting 2 pi each time...

but we wanted to solve cos(2t)=0 on (0,2pi)

so to find t divide both sides by 2
(this is a lazy way of writing 4 equations :p)

Lol. So, \[t = \frac{ \pi }{ 4 }, \frac{ 3\pi }{ 4}, ...\]

that double back Z thing is just an integer

are you still getting stuck on that? or is it clearer? or no?

I just wanted to show why I solved cos(u)=0 on [0,4pi) instead of [0,2pi)

find the points (x,y)
so we already found the t's

For example, I find that the points (-3, 0) and (3,0) are vertical tangents. ;)

k!

Well done!
http://prntscr.com/8ddra6

nice visual

@MoonlitFate do you see from @mathmate 's visual we do indeed get 4 horizontals and 2 verticals ?