<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] All my answers are correct. However, the last answer had to be obtained using a different form. I had used the general solution of $\cos^2 x = \cos ^2 \alpha$ but they expected me to use the general form of $\cos x = 0$
"Correction": 
$\cos^2 y = \frac 1 2 \implies \cos 2y = \cos 0 \implies 2y = (2n+1)\frac\pi2 \implies x = (2n+1)\dfrac \pi 8$
Any alternative methods to solve the equation are welcome. 
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