Answer
$s\in\{ \displaystyle \frac{\pi}{3}$ , $\displaystyle \frac{2\pi}{3},\ \displaystyle \frac{4\pi}{3},\ \displaystyle \frac{5\pi}{3} \}$
Work Step by Step
For any real number $s$ represented by a directed arc on the unit circle,
$\sin s=y\quad \cos s=x \quad \displaystyle \tan s=\frac{y}{x} (x\neq 0)$
Use Figure 13 on page 111.
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In the interval $[0,2\pi)$, (the unit circle), we search for
points (x,y), such that $(\displaystyle \frac{y}{x})^{2}=3$
($\displaystyle \tan s=\frac{y}{x} (x\neq 0)$
$(\displaystyle \frac{y}{x})^{2}=3\qquad /\sqrt{...}$
$\displaystyle \frac{y}{x}=\pm\sqrt{3}$
In quadrant I we find the point
$(\displaystyle \frac{1}{2},\frac{\sqrt{3}}{2})$ for which $ \displaystyle \frac{y}{x}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}$
This point is assigned to $60^{o}$, or $s=\displaystyle \frac{\pi}{3}$ rad
Applying symmetry, we find the points in the other three quadrants
(for which $\displaystyle \frac{y}{x}=\pm\sqrt{3}$):
Q.II:$\quad (-\displaystyle \frac{1}{2},\frac{\sqrt{3}}{2}),\quad\frac{y}{x}=-\sqrt{3}, \quad s=\frac{\pi}{3}$
Q.III:$\displaystyle \quad(-\frac{1}{2},-\frac{\sqrt{3}}{2}),\quad\frac{y}{x}=+\sqrt{3}, \quad s=\frac{4\pi}{3}$
Q.IV:$\displaystyle \quad(\frac{1}{2},-\frac{\sqrt{3}}{2}),\quad\frac{y}{x}=-\sqrt{3}, \quad s=\frac{5\pi}{3}$
$s\in\{ \displaystyle \frac{\pi}{3}$ , $\displaystyle \frac{2\pi}{3},\ \displaystyle \frac{4\pi}{3},\ \displaystyle \frac{5\pi}{3} \}$
