[tex]32.\quad tan\theta=-\dfrac{\sqrt3}{3}\\\\.\qquad \dfrac{sin\theta}{cos\theta}=-\dfrac{\sqrt3}{3}\\\\\text{Since there is no "3" on the Unit Circle, un-rationalize the denominator:}\\.\qquad \dfrac{sin\theta}{cos\theta}=-\dfrac{\sqrt3}{3}\bigg(\dfrac{\sqrt3}{\sqrt3}\bigg)\implies \dfrac{sin\theta}{cos\theta}=-\dfrac{3}{3\sqrt3}=-\dfrac{1}{\sqrt3}\\\\\implies sin\theta =\pm\dfrac{1}{2}\quad and \quad cos\theta = \pm\dfrac{\sqrt3}{2}\quad and \text{ coordinate is in Quadrant 2 or 4}[/tex]
[tex]\text{Look on the Unit Circle (below) for coordinates }\bigg(\dfrac{-\sqrt3}{2},\dfrac{1}{2}\bigg)\ and\ \bigg(\dfrac{\sqrt3}{2},\dfrac{-1}{2}\bigg)\\\\\bold{Answer:}\large\boxed{150^o\ and\ 330^o\implies \dfrac{5\pi}{6}\ and\ \dfrac{11\pi}{6}}[/tex]
[tex]33.\quad cos^2\theta=\dfrac{1}{2}\\\\\\.\qquad \sqrt{cos^2\theta}=\sqrt{\dfrac{1}{2}}\\\\\\.\qquad cos\theta=\pm\dfrac{1}{\sqrt2}\\\\\\\text{Rationalize the denominator:}\\.\qquad cos\theta=\pm\dfrac{1}{\sqrt2}\bigg(\dfrac{\sqrt2}{\sqrt2}\bigg)\implies cos\theta=\pm\dfrac{\sqrt2}{2}\\\\\\\text{Look on the Unit Circle to find when }cos\theta=\dfrac{\sqrt2}{2}\ and\ \dfrac{-\sqrt2}{2}[/tex]
[tex]\bold{Answer:}\large\boxed{45^o, 135^o, 225^o, 315^o\implies \dfrac{\pi}{4}, \dfrac{3\pi}{4}, \dfrac{5\pi}{4}, \dfrac{7\pi}{4}}[/tex]
[tex]34.\quad sec\theta=\text{unde fined}\implies sec\theta=\dfrac{1}{0}\\\\\\.\qquad \dfrac{1}{cos\theta}=\dfrac{1}{0}\implies cos\theta=0\\\\\\\text{Look on the Unit Circle to find when }cos\theta=0\\\\\bold{Answer:}\large\boxed{90^o\ and\ 270^o\implies\dfrac{\pi}{2}\ and\ \dfrac{3\pi}{2}}[/tex]
