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I have the following sin equation which I am supposed to graph: sin(3x) = -1 and also find how many solutions it contains between 0 and . Seeing this I am a bit confused as I don't understand what I am supposed to do with the -1. Usually when I graphed these they just equaled y. This said this is what I tried:

A (amplitude): 1

B (the stretch/compression amount): 3

P (the period): $\frac{2\pi}{3}$

Note: I found the period by doing $\frac{2\pi}{B}$

Should I bring the -1 on the other side of the equation and let it act as the axis of the wave / mid-line?

Basically my question is: what should I do with the -1? How does it affect the graph?

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    $\begingroup$ Just graph $y=\sin(3x)$, then find the points where the $y$-coordinate is $-1$. The problem was badly stated $\endgroup$ – Edward Jiang Jan 4 '15 at 2:21
  • $\begingroup$ The edit made to the question kind of changes things... $\endgroup$ – turkeyhundt Jan 4 '15 at 2:23
  • $\begingroup$ @turkeyhundt the edit completely changed my problem. I am not sure why it was made. $\endgroup$ – Cristian Gutu Jan 4 '15 at 2:24
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You can think of it also as, Where do $y=\sin{(3x)}$ and $y=-1$ intersect.

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You should graph $y = \sin(3x) + 1$ and look at the points where $y = 0$, i.e. the points where the graph crosses the y-axis.

Edit: As Edward Jiang pointed out in the comments, the above is technically incorrect. You are asked to graph $\sin(3x) = -1$. Hence, to find the values of $x$ that solve this equation, just graph $y = \sin(3x)$ and find the points where the graph crosses the line $y = -1$. However, what I wrote above will give the same answers.

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  • $\begingroup$ Technically, that's the wrong graph... $\endgroup$ – Edward Jiang Jan 4 '15 at 2:23
  • $\begingroup$ How so? Is it not equivalent? $\endgroup$ – Mark Jan 4 '15 at 2:28
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    $\begingroup$ Oh, I misread the question. Sorry! However, this will give equivalent values for $x$. $\endgroup$ – Mark Jan 4 '15 at 2:28
  • $\begingroup$ @EdwardJiang I had this same idea in my mind. Why is this wrong? $\endgroup$ – Cristian Gutu Jan 4 '15 at 2:29
  • $\begingroup$ The graph is shifted one unit up. It gives equivalent values of $x$, but the graph is wrong. $\endgroup$ – Edward Jiang Jan 4 '15 at 2:30

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