Answer
a) $\theta=\dfrac{8\pi}{3}+4 \pi k$
b) No solution
Work Step by Step
a) $\tan \dfrac{\theta}{4}+\sqrt 3=0$
or, $\dfrac{\theta}{4}=-\sqrt 3$
This gives: $\dfrac{\theta}{4}=\dfrac{2\pi}{3}$
Need to add $\pi k$ in $\dfrac{\theta}{4}$ because the time period for $\tan$ is $\pi$.
Thus, $\dfrac{\theta}{4}=\dfrac{2\pi}{3}+\pi k$
or, $\theta=\dfrac{8\pi}{3}+4 \pi k$
b) In order to get the solutions in the interval $[0,2 \pi)$, we will have to put $k=0$, then we get $\theta=\dfrac{8\pi}{3}$, which is greater than $2 pi$
Thus, there is no solution for the interval $[0,2 \pi)$.
