If $0\le x<2\pi \text{, }$then the number of real value of $x$ which satisfy the equation $\cos x+\cos 2x+\cos 3x+\cos 4x=0,\text{ is }$

$\cos x+\cos 2x+\cos 3x+\cos 4x=0$

$\begin{array}{l}\Rightarrow (\cos x+\cos 3x)+(\cos 2x+\cos 4x)=0\\ \Rightarrow 2\cos x\cos 2x+2\cos x\cos 3x=0\end{array}$

$\begin{array}{l}\Rightarrow 2\cos x(\cos 2x+\cos 3x)=0\\ \Rightarrow 2\cos x\cdot 2\cos \frac{5x}{2}\cos \frac{x}{2}=0\end{array}$

When   $x\in [0,2\pi )$ we have the solution as

$x=\frac{\pi }{2},\frac{3\pi }{2},\pi ,\frac{\pi }{5},\frac{3\pi }{5},\frac{7\pi }{5},\frac{9\pi }{5}$

Thus we have 7 solutions.