More circular functions help?

[username]

Hello, the following questions are from Methods checkpoints and I need a bit of help with them.

For the equation sin(2x) = 1, the sum of the solutions in the interval [0, 4pi] is:

Here is what I've done:

sin(2x) = 1 =
--> It's in the first and second quadrant so the first two solutions are
--> I added/subtracted 2pi from these two answers to get   and so on.
But in the answers, I did this step wrong. They only got:

2. The number of solutions of the equation .5cos(2x) = 1 for [-pi, pi] is?

Thank you! 

[ell]

(since 1 is positive, we ignore etc)

Given domain is

Multiply the given domain by 2 (since - for example, if it was we would multiply the given domain by 5) to see how many solutions we need to go up to:





   ( is the highest we can go, goes over )


   (dividing all solutions by 2)

[username]

I still don't understand why we ignore it. I thought sin was positive in the 1st and 2nd quadrant?

Like in the unit circle:

S  A
T  C

I'm so confused, but thank you.

[enwiabe]

Because sin(pi/2) is on the border of the first and second quadrants. sin(3pi/2) is on the border of the 3rd and 4th quadrant where it would be negative.

[bucket]

but 1 is not in any of those quadrants!

edit*
yeah what he said ^^^^ lol

[Mao]

remember that 1 and -1 are the peaks, therefore [in our graph], there is only 1 solution for either of them

as for all other points, there are two each.

[username]

Oh! I understand now.

Thank you!

[Mao]

oh and, seeing your method to get the symmetrical angles, they arent exactly correct.

to get from [quadrant to quadrant]:

first to second:

first to third:

first to fourth:

second to third:

second to fourth:

third to fourth:

you should graphically imagine these, using the unit circle is always the best thing to do.
just remember, by symmetry, the angle from a line to the X axis is the angle that remains for all four quadrants.

[username]

>_< Oops. What was I doing wrong? I thought I followed those steps like in the methods text book.. Y'know, the one with the giant quadrant and circle thing meshed together.

Thanks!

[username]

I drew a picture like the one in my book:



The way I was taught in school is:

1. Establish whether its a positive or negative solution
2. Find the pi/whatever value
3. Look at the quadrant and find corresponding area that matches the positive or negative
4. Apply the neon green rule

Is that wrong? >_<

[Mao]

that is correct, though I was referring to this:

--> It's in the first and second quadrant so the first two solutions are

which, even assuming you meant , I still cannot see how you got that

[username]

Uh woops. Sorry, I kinda copy and pasted the codes so I wouldn't have to type them out again and I guess I forgot to change the second one. ^_^;;