# Is my answer to this trig question correct

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

$x = \frac{11}{12}\pi+k\pi$ 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! – Mercy King Sep 15 '12 at 12:49
So am I right or wrong? And there is a difference between my answer and the correction model's answer right? – JohnPhteven Sep 15 '12 at 12:55
Can't you tell? – Mercy King Sep 15 '12 at 13:00
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 puzzled – JohnPhteven Sep 15 '12 at 13:04

Since $$-\frac{1}{12}=\frac{11}{12}-1$$ you found exactly the same solutions.
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 $1\dfrac {11}{12}\pi$ – JohnPhteven Sep 15 '12 at 12:36
The cosine is periodic with period $2\pi$, but the square of the cosine is periodic with period $\pi$, so you're OK. – Gerry Myerson Sep 15 '12 at 13:09
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? – JohnPhteven Sep 15 '12 at 13:15
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. – Gerry Myerson Sep 16 '12 at 12:07