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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 $

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    $x=-\frac{\pi}{12}+k\pi=\frac{12-1}{12}\pi+n\pi=\frac{11}{12}\pi+n\pi$ where $n=k-1$ is again an integer!2012-09-15
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    So am I right or wrong? And there is a difference between my answer and the correction model's answer right?2012-09-15
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    Can't you tell?2012-09-15
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    I usually have trouble interpreting answers like you gave me, I am just puzzled and if you read my reaction to siminore below you will understand why I'm extremely puzzled2012-09-15

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Since $$ -\frac{1}{12}=\frac{11}{12}-1 $$ you found exactly the same solutions.

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    Yes but my teacher said it was no problem if you're answer was like $2\pi$ more/less than the answer since a circle is $2\pi$ radians so aren't these different answers? I have $ \dfrac {11}{12}\pi$ instead of $ 1\dfrac {11}{12}\pi$2012-09-15
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    The cosine is periodic with period $2\pi$, but the *square* of the cosine is periodic with period $\pi$, so you're OK.2012-09-15
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    Oh, I understand. Thank you for clarifying that! Is that so because negative integers become positive when squared, thus eliminating half of all answers= period $\pi$ ? This is just a wild guess and I probably formulated my conjecture wrongly, but I hate not knowing why something is true, especially in maths! Also, just to be certain, so both my answer and that of the correction model are 100% correct?2012-09-15
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    It has nothing to do with integers. $\cos(x+\pi)=-\cos x$; if you can prove that, then squaring both sides you get $\cos^2(x+\pi)=\cos^2x$, so square of cosine has period $\pi$. By the way, if you want to be sure I see a comment intended for me, you have to include @Gerry in it somewhere.2012-09-16