If $\sin \:2 \theta= \cos \:3\theta .$ The number of elements for the set $\theta$ in $0\leq ;\theta \leq 2\pi .$

Given,

$sin2\theta = cos3\theta$

$\Rightarrow cos3\theta = cos(\dfrac{\pi}{2} - 2\theta)$

Therefore general solution is,

$3\theta = 2n\pi \pm (\dfrac{\pi}{2} - 2\theta)$, where n is any integer

$\Rightarrow \theta = 2n\pi - \dfrac{\pi}{2}, (4n + 1)\dfrac{\pi}{10}$

thus solution in given interval is,

$\theta = \dfrac{\pi}{10}, \dfrac{\pi}{2}, \dfrac{9\pi}{10}, \dfrac{13\pi}{10}, \dfrac{3\pi}{2}, \dfrac{17\pi}{10}$

Hence, option 'C' is correct.