Trigonometry Tangents

We have

$0^{\circ}<\theta<360^{\circ}$

$\implies 0^{\circ}<2\theta<720^{\circ}$

$\implies -330^{\circ}<2\theta-330^{\circ}<390^{\circ}$

Since $\tan$ is only positive when $2\theta-330^{\circ}$ lies in the first and third quadrant, and $\tan 60^{\circ}=\sqrt{3}$ , we have $2\theta-330^{\circ}=60^{\circ}, 240^{\circ}$ .

Since $\tan$ has a periodicity of $180^{\circ}$ , the following values of $2\theta-330^{\circ}$ are possible: $-300^{\circ}, -120^{\circ}, 60^{\circ}, 240^{\circ}$ .

Thus $\theta=15^{\circ}, 105^{\circ}, 195^{\circ}, 285^{\circ}$ . Summing these up we have $\boxed{600^{\circ}}$ and we are done.

using notable angles, we know that:

$\tan (\beta) = \sqrt3$ when $\beta = 60 + 180*n$ deg, (1) w/ $n = 0,\pm1,\pm2,\pm3...$

So, making equal the trigonometric fuction argument with (1):

$\Rightarrow 2\theta -330=60+180n$

$\Rightarrow \theta = 195+90n$

which values included in $0 < \theta < 360$ deg. are:

$\Rightarrow \theta = 15,105,195,285$ and suming these numbers we obtain 600 degres

according to trigonometric equations if tan(alpha)=tan(theta) then alpha=n pi+(theta), then 2x-330=n pi+60 where n is integer, solving this for n=0,1,-1,-2 we get alpha=13 pi/12, 19 pi/12, 7 pi/12, pi/12...summing this we 10 pi/3 which is 600`