Answer
$\color{blue}{s=\dfrac{7\pi}{6}}$
Work Step by Step
RECALL:
$\sec{x} = \dfrac{1}{\cos{x}}$
Since $\sec{s} = -\dfrac{2\sqrt3}{3}$, then
$\dfrac{1}{\cos{s}}=-\dfrac{2\sqrt3}{3}$
Cross-multiply to obtain:
\begin{array}{ccc}
&1(3) &= &-2\sqrt3 \cdot \cos{s}
\\&3 &= &-2\sqrt3 \cdot \cos{s}
\\&\dfrac{3}{-2\sqrt3} &= &\dfrac{-2\sqrt3 \cdot \cos{s}}{-2\sqrt3}
\\&\dfrac{3}{-2\sqrt3} \cdot \dfrac{\sqrt3}{\sqrt3} &= &\cos{s}
\\&\dfrac{3\sqrt3}{-2(3)} &= &\cos{s}
\\&-\dfrac{\sqrt3}{2} &= &\cos{s}
\end{array}
Note that $\dfrac{\pi}{6}$ is a special angle and $\cos{(\dfrac{\pi}{6})}=\dfrac{\sqrt3}{2}$.
The value of $s$ must be in the interval $\left[\pi, \dfrac{3\pi}{2}\right]$, which is in Quadrant III.
RECALL:
An angle and its reference angle have either the same trigonometric values or they differ only in signs.
The angle $\dfrac{7\pi}{6}$ is in Quadrant III and its reference angle is $\dfrac{\pi}{6}$. Both cosine and secant are negative in Quadrrant III.
Therefore, if $\sec{s}=-\dfrac{2\sqrt3}{3}$, and $s$ is in $\left[\pi, \dfrac{3\pi}{2}\right]$, then $\color{blue}{s=\dfrac{7\pi}{6}}$.
