Trig question I don't really understand $4\cos^2 \left( x + \dfrac{1}{4}\pi \right)$ = 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..

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

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 $\,\pi\,$)