If $\cos 4\mathrm{\theta }+\sin 5\mathrm{\theta }=2\text{ then }\mathrm{\theta }=$

$\cos 4\mathrm{\theta }=1$and $\sin 5\mathrm{\theta }=1$
Thus, $4\mathrm{\theta }=2\mathrm{n\pi }\text{ and }5\mathrm{\theta }=2\mathrm{m\pi }+\mathrm{\pi }/2,\mathrm{n},\mathrm{m}\in \mathrm{Z}$
$\Rightarrow \mathrm{\theta }=\frac{2\mathrm{n\pi }}{4}\text{ and }\mathrm{\theta }=\frac{2\mathrm{m\pi }}{5}+\frac{\mathrm{\pi }}{10},\mathrm{n},\mathrm{m}\in \mathrm{Z}$
Therefore, the solution is $\mathrm{\theta }=2\mathrm{k\pi }+\mathrm{\pi }/2,\mathrm{k}\in \mathrm{Z}$