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?

Hint:
Clearly, $\sin x\ne0$
$$1=4\cos x\cos2x\cos3x$$
$$\implies\sin x=2(\sin2x)\cos2x\cos3x=\sin4x\cos3x$$
$$\implies2\sin x=2\sin4x\cos3x=\sin7x+\sin x$$
$$\implies0=\sin7x-\sin x=2\sin3x\cos4x$$
