<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] Expanding the equation in $t$ gives
$$32 t^6-40 t^4+12 t^2-1=0$$ Let $y=t^2$ to make
$$32 y^3-40 y^2+12y-1=0$$ By inspection $y=\frac 14$ is a solution; so,
$$\left(y-\frac 14 \right)(8y^2-8y+1)=0$$
I am sure that you will easily finish.
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