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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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    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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    Thanks @MichaelHardy, it certainly looks better that way.2012-09-16