The number of solution of $\tan \left(\frac{\mathrm{\pi }}{4}-\mathrm{x}\right)=1+\sin 2\mathrm{x}$ in the interval $[0,4\mathrm{\pi }]$equals______

$\begin{array}{l}\tan \left(\frac{\mathrm{x}}{4}-\mathrm{x}\right)=1+\sin 2\mathrm{x}\\ \Rightarrow \frac{1-\tan \mathrm{x}}{1+\tan \mathrm{x}}=1+\sin 2\mathrm{x}\\ \Rightarrow \cos \mathrm{x}-\sin \mathrm{x}-(\cos \mathrm{x}+\sin \mathrm{x})(1+2\sin \mathrm{xcos}\mathrm{x})\\ \Rightarrow \sin \mathrm{x}\left(2+2{\cos }^2\mathrm{x}-2\sin \mathrm{xcos}\mathrm{x}\right)=0\\ \Rightarrow \sin \mathrm{x}(3+\cos 2\mathrm{x}+\sin 2\mathrm{x})=0\\ \text{ As }3+\cos 2\mathrm{x}+\sin 2\mathrm{x}\ne 0\\ \Rightarrow \sin \mathrm{x}=0\Rightarrow \mathrm{x}=\mathrm{n\pi },\mathrm{n}\in \mathrm{\mathbb{Z}}\end{array}$
Hence the number of solutions in the interval
$[0,4\mathrm{\pi }]$equals 5.