Answer
There are three solutions in the interval $[0^{\circ},360^{\circ})$:
$\theta = 70.5^{\circ}, 180^{\circ}, 289.5^{\circ}$
Work Step by Step
$3~cos^2~\theta+2~cos~\theta-1 = 0$
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 $[0^{\circ},360^{\circ})$:
$\theta = 70.5^{\circ}, 180^{\circ}, 289.5^{\circ}$
