# Finding where x = 0

It's been a long time since I had to try and figure this stuff out, so a little refresher would be helpful.

I'm trying to find where x = 0 in this equation: sin((x-10)^2)

I forget how this works. Can I get a little help?

Thank you!

SteamKing
Staff Emeritus
Homework Helper
It's been a long time since I had to try and figure this stuff out, so a little refresher would be helpful.

I'm trying to find where x = 0 in this equation: sin((x-10)^2)

I forget how this works. Can I get a little help?

Thank you!
You have an expression, not an equation.

Are you instead trying to find out the values of x which make sin ((x-10)2) = 0 ?

Right, yes! That!

phion
Gold Member
I believe you mean to find where $sin((x-10)^2)) = 0$,where $x^2-20x+100=n\pi$.

Mark44
Mentor
I believe you mean to find where $sin((x-10)^2)) = 0$,where $x^2-20x+100=n\pi$.
There's no advantage in expanding (x - 10)2 in the above. Just solve the equation ##(x - 10)^2 = n\pi## for x.

So, x = (√(nπ) + 10)? So then where does arcsin come in?

SteamKing
Staff Emeritus
Homework Helper
So, x = (√(nπ) + 10)? So then where does arcsin come in?
sin ((x- 10)2) = 0

arcsin [sin ((x-102)] = arcsin (0)

(x - 10)2 = nπ, where n = 0, 1, 2, 3, ...

x - 10 = √(nπ)

x = √(nπ) + 10

2thumbsGuy
Mark44
Mentor
sin ((x- 10)2) = 0

arcsin [sin ((x-102)] = arcsin (0)

(x - 10)2 = nπ, where n = 0, 1, 2, 3, ...

x - 10 = √(nπ)
Make that ##x - 10 = \pm\sqrt{n\pi}## and I'll be happy.
SteamKing said:
x = √(nπ) + 10

2thumbsGuy
aaaaaaah nice, it's kind of coming trickling back, now. Thank you, sirs!

OK, I'm confused again. So...

arcsin [sin ((x-102)] = arcsin (0) = (x - 10)2 = nπ ...?

How does arcsin(0) equal anything but 0?

Mark44
Mentor
OK, I'm confused again. So...

arcsin [sin ((x-102)] = arcsin (0) = (x - 10)2 = nπ ...?

How does arcsin(0) equal anything but 0?
It doesn't. In my opinion you'll just confuse yourself by invoking arcsin.

You're trying to solve the equation ##\sin((x - 10)^2) = 0##. The easiest way to solve this is to recognize that sin(A) = 0 when A is any integer multiple of ##\pi##, so the solution of this equation is ##A = k\pi##, where k is any integer. Look at a graph of y = sin(x) to see this.

Going back to your equation, you have ##\sin((x - 10)^2) = 0##, so it must be that ##(x - 10)^2 = k\pi##, with k again being any integer. Can you solve this equation for x now?

2thumbsGuy
OK, that helps. I've actually been working with a much more complex expression that turns out I would need to use imaginary numbers to fully plot, so that hasn't helped. And also I've been mapping a series of formulas and writing checks into excel, which doesn't throw zero on sin(pi), so I'm getting some red herrings. But once I've accounted for these things and done them manually or with simpler functions I've had more luck.

Many thanks!