## Trigonometry (10th Edition)

There are three solutions in the interval $[0^{\circ},360^{\circ})$: $\theta = 70.5^{\circ}, 180^{\circ}, 289.5^{\circ}$
$3~cos^2~\theta+2~cos~\theta-1 = 0$ We can use the quadratic formula to find the solutions for $cos~\theta$: $cos~\theta = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ $cos~\theta = \frac{-2 \pm \sqrt{2^2-(4)(3)(-1)}}{(2)(3)}$ $cos~\theta = \frac{-2 \pm \sqrt{16}}{6}$ $cos~\theta = \frac{1}{3}, -1$ When $cos~\theta = \frac{1}{3}$: $\theta = 70.5^{\circ}, 289.5^{\circ}$ When $cos~\theta = -1$: $\theta = 180^{\circ}$ There are three solutions in the interval $[0^{\circ},360^{\circ})$: $\theta = 70.5^{\circ}, 180^{\circ}, 289.5^{\circ}$