$\cos x+\cos3x+2\cos5x=0\\⇔(\cos x+\ \cos5x)+(\cos 5x+\cos3x)=0\\⇔\cos3x\cos2x+\cos4x\cos x=0\\⇔(4\cos^3x-3\cos x)\cos2x+\cos4x\cos x=0\\⇔\cos x[(4\cos^2x-3)\cos2x+\cos4x]=0\\\begin{cases}\cos x=0\\(2\cos2x-1)\cos2x+2\cos^2 2x-1=0\end{cases}\\⇔\begin{cases}x=\dfrac{π}{2}+kπ\\4\cos^2 2x-\cos2x-1=0\end{cases}\\⇔\begin{cases}x=\dfrac{π}{2}+kπ\\\cos2x=\dfrac{1±\sqrt{17}}{8}\end{cases}\\⇔\begin{cases}x=\dfrac{π}{2}+kπ\\x=±\dfrac{1}{2} \arccos(\dfrac{1±\sqrt{17}}{8})+kπ\end{cases}(k\in\mathbb{Z})$
