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So I got this trig equation $\cos \frac{x}{2}=1$. In theory, it's pretty simple. The problem I ran into was that the solution was outside the specified domain.

I started by finding angles with $\cos=1$. The only one was $2\pi$. So, I set up $\frac{x}{2}=2\pi$. Isolating $x$, that became $x=4\pi$.

But, the question asks for solutions between $[0,2\pi]$. I realize that $2\pi$ and $4\pi$ are coterminal angles, but does that mean that $2\pi$ is also a solution to this equation?

Another thing I was a little unsure of was whether the angle with a cosine of 1 should be considered $0$, $2\pi$, or both.

Thanks!

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    $\begingroup$ Hint: what is $\cos 0$? $\endgroup$ – 2012ssohn Jan 19 '14 at 20:03
  • $\begingroup$ $cos0=1$. So is $0$ the solution here? Part of what I'm confused about is that $cos2\pi=1$ as well. $\endgroup$ – evamvid Jan 19 '14 at 20:04
  • $\begingroup$ I agree with the above hint. I also wanted to add that since your interval contains both o and $2\pi$, then cosine is equal to one at both angles. $\endgroup$ – monroej Jan 19 '14 at 20:04
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    $\begingroup$ Actually, because it's $x$ that ranges from $0$ to $2\pi$, the angle would range from $0$ to $\pi$. $\endgroup$ – 2012ssohn Jan 19 '14 at 20:05
  • $\begingroup$ Right, but my problem is that when you multiply both sides by 2, the resulting solution is out of the equation. The idea I'm starting to get from you and 2012ssohn is that $0$ is a solution but $2\pi$ is not. Is that correct? $\endgroup$ – evamvid Jan 19 '14 at 20:06
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The solutions are $\frac{x}{2}=2\pi$ for $k=0,\pm1,\pm2,\ldots$ thas is $x=4k\pi$ There is no solution in $[0,2\pi]$ except $x=0$

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Hint:

$\cos y=\cos\alpha$ gives $y=\pm\alpha+2k\pi$ for $k\in\mathbb{Z}$

Here you can take $y=\frac{x}{2}$ and $\alpha=0$.

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  • $\begingroup$ I don't understand what $k∈Z$ means. It looks like a range of values, but I'm not sure where the big z-looking thing came from:) $\endgroup$ – evamvid Jan 19 '14 at 20:12
  • $\begingroup$ It means that $k$ is an integer. $\endgroup$ – drhab Jan 19 '14 at 20:17
  • $\begingroup$ Oh...okay. Thanks! $\endgroup$ – evamvid Jan 19 '14 at 20:18
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You know that $\cos\theta = 1$ for every angle $\theta_k = 2\pi k, k = 0,\pm 1,\pm 2,\ldots$, so $x_k = 2\theta_k$ for every $k\in \mathbb{Z}$.

Cheers!

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  • $\begingroup$ I have the same problem here as below; namely, what's $k∈Z$? $\endgroup$ – evamvid Jan 19 '14 at 20:17
  • $\begingroup$ never mind; DrHab explained it. Thanks! $\endgroup$ – evamvid Jan 19 '14 at 20:18

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