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
Best Answer
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
$\implies0\le\dfrac\pi6(6n\pm1)\le2\pi\iff0\le6n\pm1\le12$
$0\le6n+1\le12\implies n=0,1$
$0\le6n-1\le12\implies n=1,2$