number of distinct solution $x\in[0,\pi]$ of the equation satisfy $8\cos x\cos 4x\cos 5x=1$

Question

The number of distinct solution $x\in[0,\pi]$ of the equation which satisfy $8\cos x\cos 4x\cos 5x=1$

Try: $$4\bigg[\cos(6x)+\cos(4x)\bigg]\cos 4x=1$$

$$2\bigg[\cos(10x)+\cos(2x)+1+\cos (8x)\bigg]=1$$

So $$\bigg[\cos(10x)+\cos(8x)+\cos(2x)\bigg]=-1$$

Could some help me to solve it, Thanks

Answer 1

Let $\cos{x}=t$.

Thus, we need to solve $$8t(8t^4-8t^2+1)(16t^5-20t^3+5t)=1$$ or $$(8t^3+4t^2-4t-1)(8t^3-4t^2-4t+1)(16t^4-16t^2+1)=0,$$ which gives $10$ roots on $[-1,1]$.

We can just solve this equation.

An interesting root: $t=\cos\frac{2\pi}{7}$ or $t=\cos15^{\circ}.$

Indeed, $$16t^4-16t^2+1=0$$ gives $$1-16\cos^2x(1-\cos^2x)=0$$ or $$1-4\sin^22x=0$$ or $$1-2(1-\cos4x)=0$$ or $$\cos4x=\frac{1}{2}$$ or $$x=\pm15^{\circ}+90^{\circ}k,$$ where $k=\mathbb Z$ and we get here four roots: $$\{15^{\circ},105^{\circ},75^{\circ},165^{\circ}\}.$$ Now, $$8t^3+4t^2-4t-1=0$$ gives $$8\cos^3x+4\cos^2x-4\cos{x}-1=0$$ or $$8\cos^3x-6\cos{x}+4\cos^2x+2\cos{x}-1=0$$ or $$2\cos3x+2(1+\cos2x)+2\cos{x}-1=0$$ or $$2(\cos{x}+\cos2x+\cos3x)=-1$$ and since $\sin\frac{x}{2}\neq0$ we obtain: $$2\sin\frac{x}{2}(\cos{x}+\cos2x+\cos3x)=-\sin\frac{x}{2}$$ or $$\sin\frac{3x}{2}-\sin\frac{x}{2}+\sin\frac{5x}{2}-\sin\frac{3x}{2}+\sin\frac{7x}{2}-\sin\frac{5x}{2}=-\sin\frac{x}{2}$$ or $$\sin\frac{7x}{2}=0$$ or $$x=\frac{360^{\circ}k}{7},$$ where $k\in\mathbb Z$, which gives $$\left\{\frac{360^{\circ}}{7},\frac{720^{\circ}}{7},\frac{1080^{\circ}}{7}\right\}.$$ The equation $8t^3-4t^2-4t+1=0$ also gives three roots.

Another way.

Since $\sin{x}\neq0,$ by your work we obtain $$2\sin{x}(\cos10x+\cos8x+\cos2x)=-\sin{x}$$ or $$\sin11x-\sin9x+\sin9x-\sin7x+\sin3x-\sin{x}=-\sin{x}$$ or $$\sin11x+\sin3x-\sin7x=0$$ or $$2\sin7x\cos4x-\sin7x=0,$$ which gives $$\sin7x=0$$ or $$\cos4x=\frac{1}{2}$$ and we get the same $10$ roots.

Answer 2

Clearly, $\sin x\ne0$

and what if $\cos2x=0?$

For $\sin x\cos2x\ne0,$ $$8\cos x\cos4x=\dfrac{\sin8x}{\sin x\cos2x}$$

$$8\cos x\cos4x\cos5x=1\implies2\sin x\cos2x=2\cos5x\sin8x$$

$$\sin3x-\sin x=\sin13x+\sin3x$$

$$0=\sin13x+\sin x=2\sin7x\cos6x$$

Now if $\sin7x=0\implies7x=m\pi$ where $m$ is any integer

$\implies0\le\dfrac{m\pi}7\le\pi\iff0\le m\le7$

But $\sin x\ne0,0<m<7$ which accounts for $6$ roots

or if $\cos6x=0$ But $\cos2x\ne0$

$$\implies\dfrac{\cos6x}{\cos2x}=4\cos^22x-3=0$$

$\implies\cos^22x=\dfrac34\implies2x=n\pi\pm\dfrac\pi6=\dfrac\pi6(6n\pm1)$ where $n$ is any integer

$\implies0\le\dfrac\pi6(6n\pm1)\le2\pi\iff0\le6n\pm1\le12$

$0\le6n+1\le12\implies n=0,1$

$0\le6n-1\le12\implies n=1,2$