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

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.
