- #1

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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!

- Thread starter 2thumbsGuy
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- #1

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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!

- #2

SteamKing

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You have an expression, not an equation.

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!

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

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Right, yes! That!

- #4

phion

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- #5

Mark44

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There's no advantage in expanding (x - 10)

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

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SteamKing

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

arcsin [sin ((x-10

(x - 10)

x - 10 = √(nπ)

x = √(nπ) + 10

How about that?

- #8

Mark44

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Make that ##x - 10 = \pm\sqrt{n\pi}## and I'll be happy.sin ((x- 10)^{2}) = 0

arcsin [sin ((x-10^{2})] = arcsin (0)

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

x - 10 = √(nπ)

SteamKing said:x = √(nπ) + 10

How about that?

- #9

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

- #10

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arcsin [sin ((x-10

How does arcsin(0) equal anything but 0?

- #11

Mark44

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It doesn't. In my opinion you'll just confuse yourself by invoking arcsin.

arcsin [sin ((x-10^{2})] = arcsin (0) = (x - 10)^{2}= nπ ...?

How does arcsin(0) equal anything but 0?

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?

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Many thanks!