The number of roots of the equation $\cos 3x+\cos 2x=\sin \frac{3x}{2}+\sin \frac{x}{2}$in $[0,2\pi ]$ is

$\cos 3x+\cos 2x=\sin \frac{3x}{2}+\sin \frac{x}{2}$

$\begin{array}{l}\Rightarrow 2\cos \frac{5x}{2}\cos \frac{x}{2}=2\sin x\cos \frac{x}{2}\\ \Rightarrow \cos \frac{x}{2}\left(\cos \frac{5x}{2}-\sin x\right)=0\\ \Rightarrow \cos \frac{x}{2}=0\Rightarrow \frac{x}{2}=(2n+1)\frac{\pi }{2}\\ \Rightarrow x=(2n+1)\pi \Rightarrow x=\pi \text{ in }[0,2\pi ]\end{array}$

And, $\cos \frac{5x}{2}=\sin x=\cos \left(\frac{\pi }{2}-x\right)$

$\begin{array}{l}\Rightarrow \frac{5x}{2}=2n\pi \pm \left(\frac{\pi }{2}-x\right)\\ \Rightarrow \frac{5x}{2}=2n\pi +\frac{\pi }{2}-x\Rightarrow \frac{7x}{2}=(4n+1)\frac{\pi }{2}\\ \Rightarrow x=(4n+1)\frac{\pi }{7}\Rightarrow x=\frac{\pi }{7},\frac{5\pi }{7},\frac{9\pi }{7},\frac{13\pi }{7}\end{array}$

$\Rightarrow$ $\frac{5x}{2}=2n\pi -\frac{\pi }{2}+x\Rightarrow \frac{3x}{2}=(4n-1)\frac{\pi }{2}$

$\Rightarrow x=(4n-1)\frac{\pi }{3}\Rightarrow x=\pi$

$\therefore$The roots are $\frac{\pi }{7},\frac{5\pi }{7},\pi ,\frac{9\pi }{7},\frac{13\pi }{7}$.