Answer
a) $\theta=4 \pi+6 \pi k$ or, $\theta=5 \pi+6 \pi k$
b) No solution
Work Step by Step
a) $2 \sin \dfrac{\theta}{3}+\sqrt 3=0$
or, $\sin \dfrac{\theta}{3}=-\dfrac{\sqrt 3}{2}$
This gives: $\dfrac{\theta}{3}=\dfrac{4\pi}{3}$
Need to add $2\pi k$ in $\dfrac{\theta}{3}$ because the time period for $\sin$ is $2 \pi$.
Thus, $\dfrac{\theta}{3}=\dfrac{4\pi}{3}+2\pi k$ or, $\dfrac{\theta}{3}=\dfrac{5\pi}{3}+2\pi k$
so, $\theta=4 \pi+6 \pi k$ or, $\theta=5 \pi+6 \pi k$
b) In order to get the solutions in the interval $[0,2 \pi)$, we will have to put $k=0$, then we get $\theta=4 \pi, 5 \pi$, and $k=-1$, then we get $\theta=-2 \pi, - \pi$,
Thus, there is no solution for the interval $[0,2 \pi)$.
