Strange Trig Function solutions

[WolfOfTheSteps]

Homework Statement

This is part of a problem for a nonlinear diff class... But it's the basic stuff that's tripping me up.

Find all the max/min and concavity for

[tex]v(x) = -cos(x)-Lx+1,\ \ \ 0<L<1 [/tex]

The Attempt at a Solution

Here's what I do:

[tex] v'(x) = sin(x)-L [/tex]
[tex]v''(x)=cos(x)[/tex]

Set the first derivative to 0:

[tex]sin(x)-L=0 \Rightarrow x = arcsin(L)[/tex]

Here's where I'm confused. I say

[tex]x = arcsin(L) + 2\pi n, \ \ \ \mbox{where }n \mbox{ is any integer}[/tex]

But Maple says:

[tex]x = arcsin(L)+2\pi n, \ \ \ \mbox{where }n \mbox{ is any integer, OR:}[/tex]
[tex]x = arcsin(L) - 2arcsin(L)+2\pi n+\pi \ \ \ \mbox{where }n \mbox{ is any integer}[/tex]

Where does the -2arcsin(L) and the +pi come from?

I really want to understand this once and for all... I never took a trig class, and while I get by fine 99% of the time, I hit a brick wall when I come across this kind of stuff.

Thanks!

 

[WolfOfTheSteps]

I think I almost understand it... The second set of values of x are the ones where the arcsin is in the II quadrant. Correct?

And obviously it makes more sense to write:

[tex] x = \pi - arcsin(L) + 2\pi n[/tex]

for the 2nd set of x values.

Yeah, I think I get it now. I was just looking at the graph of arcsine, and it doesn't seem to make sense from that.

Is there a way to see that those are the solutions by looking at the arcsine graph?

Thanks.

 

[arunbg]

I hope you understand the first solution.
The second part of the solution is better understood if you manipulate the terms a bit , it can be written as [itex]x = (2n+1)\pi-arcsin(L)[/itex].
Noting that [itex]sin(n\pi-x)=sin(x) [/itex] for all odd n, you can now see how this is also part of the solution.

If you still don't follow, perhaps substitution of values for L into the solutions can give a clearer picture.
For the graphical picture, you should use the sin(x) graph. Fix some L value, find arcsin(L), and see what other values of x, do you get sin(x)=L, that's all.