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

$\cos 4\theta =1\text{ and }\sin 5\theta =1\text{, }$

$\text{ Thus, }4\theta =2n\pi \text{ and }5\theta =2m\pi +\pi /2,n,m\in Z$

$\Rightarrow \theta =\frac{2n\pi }{4}\text{ and }\theta =\frac{2m\pi }{5}+\frac{\pi }{10},n,m\in Z$

$\text{ Therefore, the solution is }\theta =2k\pi +\pi /2,k\in Z$