## Trigonometry (11th Edition) Clone

$\color{blue}{s=\dfrac{7\pi}{6}}$
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 $\color{blue}{s=\dfrac{7\pi}{6}}$.