1. ## Finding asymptotes+Fundamental identities

Hey, I'm new to calculus, and I can't figure out this problem.

y = TAN (2 SIN x) -pi < x < pi

Since: TAN x = (SIN x / COS x),

you can get

y= SIN (2 SIN x) / COS (2 SIN x)

Then, since the asymptote is where the equation is undefined (or equals infinity), we need to find the value of x which causes the denominator to equal 0.

COS (2 SIN x) = 0

The cosine of +pi/2 is 0, so

2 SIN x = +pi/2
SIN x = +pi/4
x = + SIN^-1 pi/4

The problem is that there are two answers, but I can't figure out how to work out the next one, which the book says is

x = + (pi - SIN^-1[pi/4]

Help, plz!

I'd also like a way to type in a pi symbol, if that's possible.

2. If you have an equation $\sin x = C$ then the possible solutions are $x=\arcsin C+2n\pi$ and $x=\pi - \arcsin C+2n\pi$

3. Is that just a general rule? My Math book doesn't show it. Also, does it work for just $pi - arcsin C$? Otherwise I don't see how it works. The $2n pi$ doesn't seem to fit at all. If it does, then thanks a lot.

4. It's the general solution to an equation like this. It contains all the solutions to the equation, but for your problem the question makes it clear that we are only interested in the domain $x\in [-\pi,\pi]$, so the only value of $n$ that doesn't take us out from the valid domain is $n=0$.

Hence the two solutions are $x=\pm\arcsin\frac{\pi}{4}$ and $x=\pm\left(\pi-\arcsin\frac{\pi}{4}\right)$

If you think about it, going $2n\pi$ radians in a circle always takes us back to where we started.

5. Oh, yeah, I guess I was being pretty dumb about the $2n pi$ thing. Thanks again.