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From a book I know that the roots of $\cos(x)$ are: $$\left[\ldots,-\frac{3\pi}{2},-\frac{\pi}{2},\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2},\ldots\right]$$

What are the roots of $2\cos(x)$?

I have the answer here: $$\left[\ldots,-\frac{\pi}{2},\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2},\ldots\right]$$

I just don't see the connection or how this was arrived at.

Why are those the roots of $2\cos(x)$?

Robert Howard
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Doug Fir
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2 Answers2

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If $x$ is a real number such that $\cos(x) = 0$, what is the value of $2\cos(x)$?

On the other hand, if $x$ is a real number such that $2\cos(x) = 0$, what is the value of $\cos(x)$?

Now can you see why the roots of $\cos(x)$ and the roots of $2\cos(x)$ are the same?

Arthur
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  • Hi Arthur thanks for your answer. Are you saying that because cos(x) = 0 then 2cos(x) must also be 0 and thus they are the same? I'm confused because the two screen shots in my post are different e.g. The first item in the first screen shot from my book showing the roots of cos(x) is 3pi/2 whereas the first item in the second screen shot which is supposed to be the roots of 2cos(x) is just pi/2? – Doug Fir Sep 13 '18 at 14:54
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    @DougFir The $\cdots$ in the two lists make them the same, even though they're not written down in entirely identical ways. There is no "first item", just "arbitrary item somewhere in the middle of the list showing how the list works". – Arthur Sep 13 '18 at 15:16
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Hint:

If $2\cdot \cos(x)=0$, then either:

  • $2=0$
  • $\cos(x)=0$
cansomeonehelpmeout
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