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?
 A: It's $$\cos2x(\cos2x+\cos4x)=\frac{1}{2}.$$
Now, let $\cos2x=t$.
Thus, $$t(t+2t^2-1)=\frac{1}{2}$$ or
$$4t^3+2t^2-2t-1=0$$ or
$$2t^2(2t+1)-(2t+1)=0$$ or
$$(2t^2-1)(2t+1)=0,$$ which gives
$$\cos4x=0$$ or $$\cos2x=-\frac{1}{2}$$ and we can write the answer.
A: 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
A: Remember that a cubic equation can have up to three real solutions. You did find that $\cos y = -\dfrac{1}{2}$ is one of the solutions, but you still need to find the other two.
Since $\cos y = -\dfrac{1}{2}$ is a root of $4\cos^3 y+2\cos^2y- 2\cos y-1=0$, we can factor the equation as:
\begin{align*}
4\cos^3 y+2\cos^2y- 2\cos y-1 &= 0
\\
(2\cos y + 1)(2\cos^2 y - 1) &= 0
\\
(2\cos y + 1)(\sqrt{2}\cos y + 1)(\sqrt{2}\cos y - 1) &= 0
\end{align*}
So the solutions are $$\cos y = -\dfrac{1}{2}, -\dfrac{1}{\sqrt{2}}, \dfrac{1}{\sqrt{2}}$$
Now, solve these for $y$, and then divide by $2$ to get $x$. 
A: 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. 
