Answer
a) $\theta=2 \pi k$
b) $\theta =0$
Work Step by Step
a) $\sec \dfrac{\theta}{2}=\cos \dfrac{\theta}{2}$
or, $\cos^2 \dfrac{\theta}{2}=1$
This gives: $\cos \dfrac{\theta}{2}=\pm 1$
Need to add $2\pi k$ in $\dfrac{\theta}{2}$ because the time period for $\sin$ is $2 \pi$.
Thus, $\dfrac{\theta}{2}=\pi+2\pi k$ or, $\dfrac{\theta}{2}=2\pi k$
so, $\theta=2 \pi k$
b) In order to get the solutions in the interval $[0,2 \pi)$, we will have to put $k=0$, then we get $\theta \gt 2 \pi$
Thus, $\theta =0$.
