<s>[INST] solve for $(\cos x)(\cos2x)(\cos3x)=\frac{1}{4}$ solve for $(\cos x)(\cos2x)(\cos3x)=\frac{1}{4}$. what is the general solution for $x$.
I wrote $\cos2x$ as $2\cos^2x-1$ and $\cos3x$ as $4\cos^3x-3\cos x$. Then the expression gets reduced to $\cos x(2\cos^2x-1)(4\cos^3x-3\cos x)=\frac{1}{4}$. substituted $\cos x=t$ and got
$t(2t^2-1)(4t^3-3t)=\frac{1}{4}$. how do i proceed further?
 [/INST] $$\cos x\cdot \cos 2x\cdot\cos 3x=1/4$$
$$\implies (2\cos x\cdot\cos 3x)(2\cos 2x)=1$$
$$\implies (\cos 4x+\cos 2x)(2\cos 2x)=1$$
$$\implies 2\cos 4x\cdot\cos 2x+(2\cos^22x-1)=0$$
$$\implies 2\cos 4x\cdot\cos 2x+\cos 4x=0$$
$$\implies \cos 4x~(2\cos 2x+1)=0$$
Either, $~\cos 4x=0\implies 4x=(2n+1)\dfrac π2\implies x=(2n+1)\dfrac π8~$, for any integer $~n~$.
Or, $~2\cos 2x+1=0\implies 2\cos 2x=-1 \implies \cos 2x=-\dfrac 12 \implies  \cos 2x=\cos\left(π-\dfrac π3\right)~$
$~\implies  \cos 2x=\cos \left(\dfrac {2π}3\right) \implies  2x=2nπ\pm \dfrac {2π}3 \implies x=nπ\pm\dfrac π3 ~$, for any integer $~n~$.
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