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?
 A: $$\cos x\cdot \cos 2x\cdot\cos 3x=1/4$$
$$\implies (2\cos x\cdot\cos 3x)(2\cos 2x)=1$$
$$\implies (\cos 4x+\cos 2x)(2\cos 2x)=1$$
$$\implies 2\cos 4x\cdot\cos 2x+(2\cos^22x-1)=0$$
$$\implies 2\cos 4x\cdot\cos 2x+\cos 4x=0$$
$$\implies \cos 4x~(2\cos 2x+1)=0$$
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 $~n~$.
A: Hint:
Clearly, $\sin x\ne0$
$$1=4\cos x\cos2x\cos3x$$
$$\implies\sin x=2(\sin2x)\cos2x\cos3x=\sin4x\cos3x$$
$$\implies2\sin x=2\sin4x\cos3x=\sin7x+\sin x$$
$$\implies0=\sin7x-\sin x=2\sin3x\cos4x$$
A: 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.
A: $$4\cos x\cos3x\cos2x=2\cos2x(\cos2x+\cos4x)=2\cos2x(\cos2x+2\cos^22x-1)$$
$$\implies0=4\cos^32x+2\cos^22x-2\cos2x-1$$
$$=2\cos^22x(2\cos2x+1)-(2\cos2x+1)$$
$$=(2\cos2x+1)(2\cos^22x-1)$$
If $2\cos2x+1=0,$
$$\cos2x=-\cos\dfrac\pi3=\cos\left(\pi-\dfrac\pi3\right)$$
Else $0=2\cos^22x-1=\cos4x,4x=(2n+1)\dfrac\pi2$ where $n$ is any integer
