Answer
There are four solutions in the interval $[0^{\circ},360^{\circ})$:
$\theta = 53.5^{\circ}, 118.4^{\circ}, 233.5^{\circ}, 298.4^{\circ}$
Work Step by Step
$5~cot^2~\theta-~cot~\theta-2 = 0$
We can use the quadratic formula to find the solutions for $cot~\theta$:
$cot~\theta = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$
$cot~\theta = \frac{1 \pm \sqrt{(-1)^2-(4)(5)(-2)}}{(2)(5)}$
$cot~\theta = \frac{1 \pm \sqrt{41}}{10}$
$cot~\theta = 0.7403, -0.5403$
When $cot~\theta = 0.7403$:
$\theta = arccot(0.7403) = 53.5^{\circ}$
Another possible value is $\theta = 180^{\circ}+53.5^{\circ} = 233.5^{\circ}$
Thus: $\theta = 53.5^{\circ}, 233.5^{\circ}$
When $cot~\theta = -0.5403$:
$\theta = arccot(-0.5403)+360^{\circ}$
$\theta = -61.6^{\circ}+360^{\circ} = 298.4^{\circ}$
Another possible value is $\theta = 298.4^{\circ}-180^{\circ} = 118.4^{\circ}$
There are four solutions in the interval $[0^{\circ},360^{\circ})$:
$\theta = 53.5^{\circ}, 118.4^{\circ}, 233.5^{\circ}, 298.4^{\circ}$
