<s>[INST] 2nd solution of $\cos x \cos 2x\cos 3x= \frac 1 4 $ 
$\cos x \cos 2x\cos 3x= \dfrac 1 4 $

Attempt explained:
$(2\cos x \cos 3x)\cos 2x = \frac1 2 $
$(\cos 4x +\cos 2x )\cos 2x = \frac 1 2 \\\cos ^2y + \cos y (2\cos^2y- 1)= \frac1 2 \\ $
(Let, y = 2x)
$\implies 4\cos^3 y+2\cos^2y- 2\cos y-1=0$
I solved this equation using Rational Root Theorem and got $y = \frac 1 2$
$\implies x= m\pi \pm \dfrac\pi3 \forall x\in \mathbb {Z}$
Using Remainder theorem, the other solution is $\cos^2 y = \dfrac 1 2 $
$\implies x = \dfrac n2\pi \pm\dfrac \pi 8 $
But answer key states: $ x= m\pi \pm \dfrac\pi3or x =(2n+1)\dfrac \pi 8$
Why don't I get the second solution correct?
 [/INST] How does $\cos^2y=\dfrac12$ imply $$\dfrac{n\pi}2\pm\dfrac\pi6?$$  
In fact $\cos^2y=\dfrac12\iff\cos2y=0$
$\implies2y=m\pi+\dfrac\pi2,4y=\pi(2m+1)$ where $m$ is any integer $\ \  \  \ (1)$
Again, $\cos^2y=\cos^2\dfrac\pi4\iff\sin^2y=\sin^2\dfrac\pi4$
Using Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $
$y=n\pi\pm\dfrac\pi4,4y=\pi(4n\pm1)$ where $n$ is any integer $\ \  \  \ (2)$
Set some values of $m,n$  to identify $\{2m+1\},\{4n\pm1\}$ to be esentially the same set
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