Answer
The solution set to this problem is $$\theta\in\{38.4^\circ, 104.8^\circ,218.4^\circ,284.8^\circ\}$$
Work Step by Step
$$3\cot^2\theta-3\cot\theta-1=0$$ over interval $[0^\circ,360^\circ)$
1) Solve the equation:
$$3\cot^2\theta-3\cot\theta-1=0$$
Here we need to consider the equation a quadratic formula. In fact, if we take $\cot\theta$ as $x$, we would get
$$3x^2-3x-1=0$$
with $a=3, b=-3$ and $c=-1$
- Calculate $\Delta$: $\Delta=b^2-4ac=(-3)^2-4\times3\times(-1)=9-(-12)=21$
- Calculate $x$, or in other words, $\cot\theta$:
$$x=\cot\theta=\frac{-b\pm\sqrt\Delta}{2a}=\frac{3\pm\sqrt{21}}{6}$$
2) Apply the inverse function:
- For $\cot\theta=\frac{3+\sqrt{21}}{6}$: $$\cot\theta=\frac{3+\sqrt{21}}{6}\approx1.264$$
which means $$\theta=\cot^{-1}(1.264)$$
$$\theta\approx38.4^\circ\hspace{1cm}\text{or}\hspace{1cm}\theta\approx218.4^\circ$$
- For $\cot\theta=\frac{3-\sqrt{21}}{6}$: $$\cot\theta=\frac{3-\sqrt{21}}{6}\approx-0.264$$
which means $$\theta=\cot^{-1}(-0.264)$$
$$\theta\approx104.8^\circ\hspace{1cm}\text{or}\hspace{1cm}\theta\approx284.8^\circ$$
Therefore, $$\theta\in\{38.4^\circ, 104.8^\circ,218.4^\circ,284.8^\circ\}$$
In other words, the solution set to this problem is $$\theta\in\{38.4^\circ, 104.8^\circ,218.4^\circ,284.8^\circ\}$$
