Number of values of $x$ lying in the interval $[0, 4\mathrm{\pi }]$and satisfying the equation  $\tan 5x+\cot 3x=0$ is 

$\begin{array}{l}\tan 5x=-\cot 3x=\tan (\pi /2+3x)\\ \Rightarrow 5x=n\pi +\frac{\pi }{2}+3x,n\in \mathbf{I}\\ \Rightarrow 2x=(2n+1)\frac{\pi }{2},n\in \mathbf{I}\\ \Rightarrow x=(2n+1)\frac{\pi }{4},n\in \mathbf{I}\end{array}$ 

Now, $x\in [0,4\pi ]$

$\begin{array}{l}\Rightarrow 0\le (2n+1)\frac{\pi }{4}\le 4\pi \\ \Rightarrow -\frac{1}{2}\le n\le \frac{15}{2}\Rightarrow 0\le n\le 7\end{array}$

$\Rightarrow$ There are 8 values of $x$.