Answer
a) $\theta =\dfrac{2 k\pi}{3}+\pi$ or, $\theta =\dfrac{2k \pi}{3}$
b) $\theta=0,\dfrac{\pi}{3},\dfrac{2\pi}{3},\pi, \dfrac{4\pi}{3}$
Work Step by Step
a) $\dfrac{\sin 3 \theta}{\cos 3\theta}+1=\dfrac{1}{\cos 3\theta}$
This gives: $\cos (3\theta-\dfrac{\pi}{4})=\cos \dfrac{\pi}{4}$
Thus, we have $\theta=\dfrac{\pi}{2}, \dfrac{5\pi}{2}, \dfrac{9\pi}{2},..$
or, $\theta =\dfrac{\pi}{2}+2k \pi$ or, $\theta =2k \pi$
This implies that $\theta =\dfrac{2 k\pi}{3}+\pi$ or, $\theta =\dfrac{2k \pi}{3}$
b) From part, (a), we have $\theta =\dfrac{2 k\pi}{3}+\pi$ or, $\theta =\dfrac{2k \pi}{3}$
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 $\theta=0,\dfrac{\pi}{3},\dfrac{2\pi}{3},\pi, \dfrac{4\pi}{3}$
