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$4\cos^2 \left( x + \dfrac{1}{4}\pi \right)$ = 3

My final answer:

$ x = \frac{11}{12}\pi+k\pi $ and $x = \frac{7}{12}\pi + k\pi $

In the correction model it is $x = \frac{7}{12}\pi + k\pi $ and $x = -\frac{1}{12}\pi+k\pi$ (and $x = -\frac{1}{12}\pi+k\pi$ equals $x = 1\frac{11}{12}\pi+k\pi$ and not $ x = \frac{11}{12}\pi+k\pi $

  • I reposted this because the answers on the original question didn't suffice. Also, reposting on this forum is just like bumping your old post up right? If not, I'm sorry, I don't want to spam, but from previous times I learned that reposting only bumps up the original post..
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    When you say "(and $x = -\frac{1}{12}\pi+k\pi$ equals $x = 1\frac{11}{12}\pi+k\pi$ and not $x= \frac{11}{12}\pi+k\pi $" you're wrong. $$-\frac{1}{12}\pi+k\pi= -\frac{1}{12}\pi+\pi+k'\pi = -\frac{1}{12}\pi+\frac{12}{12}\pi+k'\pi = \frac{11}{12}\pi+k'\pi$$2012-09-15
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    Yes, but my teacher said the unit circle has a period of $2\pi$ but the difference between these answers is just $\pi$ so why are they both correct? Is that because of the initial question have $cos^2$ in it instead of just $cos$?2012-09-15
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    Your 'k' can be any integer, right? Just for fun, call this parameter 'k+1' instead. You'll notice that $-\frac{1}{12}\pi + k\pi$ becomes $-\frac{1}{12}\pi + (k+1)\pi$, which is $\frac{11}{12}\pi + k\pi$. It's nothing to do with the period of $2\pi$.2012-09-15
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    @ZafarS: Yes, it is connected with the $\cos^2$, because in general $\cos(\pi+t)=-\cos t$.2012-09-15
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    @ Andre thank you, I get it now.2012-09-15
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    @ZafarS: I notice that Gerry Myerson made a comment equivalent to mine in your earlier similar question.2012-09-15
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    @ZafarS : You have "and" where you should have "or".2012-09-16
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    I hope that your teacher said not that the *circle* has a period of $2\pi$, but that the *sine and cosine functions* have that period. But as you suspect, the function $\cos^2$ has a period of just $\pi$.2012-09-16
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    And why is that? I've heard that before on this forum, but I want to know why. I suspect it is like that because all negative integers, when squared, become positive integers, thus eliminating half, making $2\pi$ just simply $\pi$2012-09-16

1 Answers 1

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$$4\cos^2\left(x+\frac{\pi}{4}\right)=3\Longleftrightarrow \cos\left(x+\frac{\pi}{4}\right)=\pm\frac{\sqrt 3}{2}$$

And from here:

$$(1)\ (\text{With }+)\;\;\;x+\frac{\pi}{4}=\pm\frac{\pi}{6}+2k\pi\Longrightarrow x=\left\{\begin{array}-\;\;\;\;-\frac{\pi}{12}+2k\pi\\{}\\\;\;\;\;-\frac{5\pi}{12}+2k\pi\end{array}\right.\;\;\;,\,\,k\in\Bbb Z$$

$$(2)\ (\text{With }-)\;\;\;x+\frac{\pi}{4}=\pm\frac{5\pi}{6}+2k\pi\Longrightarrow x=\left\{\begin{array}-\;\;\;\;\;\;\;\;\;\frac{7\pi}{12}+2k\pi\\{}\\\;\;\;\;-\frac{13\pi}{12}+2k\pi\end{array}\right.\;\;\;,\,\,k\in\Bbb Z$$

Now observe that the second option in (1) and the first one in (1) differ by $\,\pi\,$ (up to a multiple of $\,2\pi\,$ , of course), and the same goes for the first option in (1) and the second one in (2), and from here you get the answers as you wrote them (i.e., up to multiples of $\,\pi\,$)

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    Where this answer said "$\text{With -}$", I changed it to "$\text{With }-$".2012-09-16
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    Thanks @MichaelHardy, it certainly looks better that way.2012-09-16