Answer
a) $\theta =k \pi$
and $\theta =\dfrac{\pi}{2}+2k \pi$
b) $\theta=0, \pi, 2\pi$ or, $\theta =\dfrac{\pi}{2}$
Work Step by Step
a) $1-\sin 2 \theta= \cos 2\theta$
or, $\sin \theta(\sin \theta-1)=0$
This gives: Either $\sin \theta=0$
Thus, we have $\theta=0, \pi, 2\pi, ....k \pi$ or, $\theta =k \pi$
or, $(\sin \theta-1)=0 \implies \sin \theta=1$
Thus, we have $\theta=\dfrac{\pi}{2}, \dfrac{5\pi}{2}, \dfrac{9\pi}{2},..$
or, $\theta =\dfrac{\pi}{2}+2k \pi$
b) From part, (a), we have $\theta =k \pi$
and $\theta =\dfrac{\pi}{2}+2k \pi$
In order to get the solutions in the interval $[0,2 \pi)$, we will have to put $k=0$, because the other values will make $\theta \gt 2 \pi$
Thus, we have $\theta=0, \pi, 2\pi$ or, $\theta =\dfrac{\pi}{2}$