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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. $$-\frac{1}{12}\pi+k\pi= -\frac{1}{12}\pi+\pi+k'\pi = -\frac{1}{12}\pi+\frac{12}{12}\pi+k'\pi = \frac{11}{12}\pi+k'\pi$$ –  Arthur Sep 15 '12 at 13:50
    
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 $cos$? –  JohnPhteven Sep 15 '12 at 13:54
    
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 $2\pi$. –  Billy Sep 15 '12 at 14:19
    
@ZafarS: Yes, it is connected with the $\cos^2$, because in general $\cos(\pi+t)=-\cos t$. –  André Nicolas Sep 15 '12 at 14:39
    
@ Andre thank you, I get it now. –  JohnPhteven Sep 15 '12 at 14:39

1 Answer 1

up vote 2 down vote accepted

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

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 "$\text{With }-$". –  Michael Hardy Sep 16 '12 at 0:57
    
Thanks @MichaelHardy, it certainly looks better that way. –  DonAntonio Sep 16 '12 at 2:49
    
Thank you kind sir! –  JohnPhteven Sep 16 '12 at 9:51

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