Answer
$\theta=20^{\circ}+60^{\circ}k,40^{\circ}+60^{\circ}k$,
where k is any integer.
Work Step by Step
$(\tan3\theta)^{2}=3$
or, $(\tan3\theta)=+\sqrt 3\,\,or\,\,-\sqrt 3$
For, $\tan3\theta=+\sqrt 3$,
$3\theta=\tan^{-1}(+\sqrt 3)$,
Since, $\tan3\theta$ is positive in both quadrants 1 and 3,
$3\theta=60^{\circ} \,\,or \,\,180^{\circ}+60^{\circ}=240^{\circ},$
To find all values of $3\theta$, we can add $180^{\circ}k$ where k is any integer, since the period of tan function is $180^{\circ}$.
So, $3\theta=60^{\circ}+180^{\circ}k\,\,or \,\,240^{\circ}+180^{\circ}k,$
So, $\theta=20^{\circ}+60^{\circ}k\,\,or \,\,80^{\circ}+60^{\circ}k,$
For, $\tan3\theta=-\sqrt 3$,
$3\theta=\tan^{-1}(-\sqrt 3)$,
Since, $\tan3\theta$ is negative in both quadrants 2 and 4,
$3\theta=180^{\circ}-60^{\circ}=120^{\circ} \,\,or \,\,360^{\circ}-60^{\circ}=300^{\circ},$
To find all values of $3\theta$, we can add $180^{\circ}k$ where k is any integer, since the period of tan function is $180^{\circ}$.
So, $3\theta=120^{\circ}+180^{\circ}k\,\,or \,\,300^{\circ}+180^{\circ}k,$
So, $\theta=40^{\circ}+60^{\circ}k\,\,or \,\,100^{\circ}+60^{\circ}k,$
So, all degree solutions are
$\theta=20^{\circ}+60^{\circ}k,40^{\circ}+60^{\circ}k.$
