Answer
(a) $\theta=180^{\circ}k,60^{\circ}+360^{\circ}k\,\,or \,\,120^{\circ}+360^{\circ}k$, for integral values of k.
(b) $0^{\circ},60^{\circ},120^{\circ},180^{\circ}.$
Work Step by Step
(a)
$(\sqrt 3\tan\theta-2\sin\theta\tan\theta)=0,$
or, $\tan\theta(\sqrt 3-2\sin\theta)=0,$
either, $\tan\theta=0$ or $\sin\theta=\frac{\sqrt 3}{2},$
For, $\\tan\theta=0$,
$\theta=\tan^{-1}(0)$,
$\theta=0^{\circ} \,\,or \,\,180^{\circ},$
To find all values of $\theta$, we can add $180^{\circ}k$ where k is any integer, since the period of sine function is $180^{\circ}$.
So, the general solution is $\theta=180^{\circ}k,$ k being an integer,
For, $\sin\theta=\frac{\sqrt 3}{2}$,
$\theta=\sin^{-1}(\frac{\sqrt 3}{2})$,
Since, $\sin\theta$ is positive in both quadrants 1 and 2,
$\theta=60^{\circ} \,\,or \,\,180^{\circ}-60^{\circ}=120^{\circ},$
To find all values of $\theta$, we can add $360^{\circ}k$ where k is any integer, since the period of sine function is $360^{\circ}$.
So, $\theta=60^{\circ}+360^{\circ}k\,\,or \,\,120^{\circ}+360^{\circ}k$
So, overall solution of $\theta=180^{\circ}k,60^{\circ}+360^{\circ}k\,\,or \,\,120^{\circ}+360^{\circ}k$, for integral values of k.
(b)
Now, if $0^{\circ}\leq\theta\lt360^{\circ}$, we get the solutions by substituting k=0 and k=1.
When k=0, $\theta=0^{\circ},60^{\circ},120^{\circ},180^{\circ}$
When k=1, the only possible value of $\theta=180^{\circ}$.
Hence, the final solutions of $\theta$ are $0^{\circ},60^{\circ},120^{\circ},180^{\circ}.$
