Answer
The solution set is
$$\{33.7^\circ+180^\circ n, 135^\circ+180^\circ n, n\in Z\}$$
Work Step by Step
$$\frac{2\cot^2\theta}{\cot\theta+3}=1$$
1) Solve the equation over the interval $[0^\circ,360^\circ)$
$$\frac{2\cot^2\theta}{\cot\theta+3}=1$$
$$2\cot^2\theta=\cot\theta+3$$
$$2\cot^2\theta-\cot\theta-3=0$$
$$(2\cot^2\theta+2\cot\theta)+(-3\cot\theta-3)=0$$
$$2\cot\theta(\cot\theta+1)-3(\cot\theta+1)=0$$
$$(\cot\theta+1)(2\cot\theta-3)=0$$
$$\cot\theta=-1\hspace{1cm}\text{or}\hspace{1cm}\cot\theta=\frac{3}{2}$$
- For $\cot\theta=-1$
Over the interval $[0^\circ, 360^\circ)$, there are 2 values of $\theta$ where $\cot\theta=-1$, which are $135^\circ$ and $315^\circ$.
So, $\theta=\{135^\circ, 315^\circ\}$
- For $\cot\theta=\frac{3}{2}$. That means, $$\theta=\cot^{-1}\frac{3}{2}$$
$$\theta=\tan^{-1}\frac{2}{3}$$
$$\theta\approx33.7^\circ\hspace{1cm}\text{or}\hspace{1cm}\theta\approx213.7^\circ$$
Therefore, overall, $$\theta=\{33.7^\circ,135^\circ, 213.7^\circ, 315^\circ\}$$
2) Solve the equation for all solutions
- The integer multiples of the period of the cotangent function is $180^\circ$.
- We apply it to each solution found in step 1.
- We end up with a set like this
$$\{33.7^\circ+180^\circ n, 135^\circ+180^\circ n, 213.7^\circ+180^\circ n, 315^\circ+180^\circ n,nāZ\}$$
- However, as $33.7^\circ+180^\circ n$ and $213.7^\circ+180^\circ n$ refer to the same set of values, we only need to include one of them in the final solution set. The same thing can be said about $135^\circ+180^\circ n$ and $315^\circ+180^\circ n$.
So the solution set is
$$\{33.7^\circ+180^\circ n, 135^\circ+180^\circ n, n\in Z\}$$
