Answer
The solution set is $$\{0^\circ+360^\circ n, 180^\circ+360^\circ n, n\in Z\}$$
Work Step by Step
$$\sin\theta\cos\theta-\sin\theta=0$$
1) Solve the equation over the interval $[0^\circ,360^\circ)$
$$\sin\theta\cos\theta-\sin\theta=0$$
$$\sin\theta(\cos\theta-1)=0$$
$$\sin\theta=0\hspace{1cm}\text{or}\hspace{1cm}\cos\theta=1$$
- For $\cos\theta=1$
Over the interval $[0^\circ, 360^\circ)$, there is only one value of $\theta$ where $\cos\theta=1$, which is $0^\circ$.
- For $\sin\theta=0$
Over the interval $[0^\circ, 360^\circ)$, there are 2 values of $\theta$ where $\sin\theta=0$, which are $0^\circ$ and $180^\circ$.
Therefore, overall, $$\theta=\{0^\circ, 180^\circ\}$$
(though $\theta=0^\circ$ appears in both cases, we only need to mention one time, since they are eventually the same point in the unit circle)
2) Solve the equation for all solutions
- The integer multiples of the period of both the sine and cosine function is $360^\circ$.
- We apply it to each solution found in step 1.
So the solution set is $$\{0^\circ+360^\circ n, 180^\circ+360^\circ n, n\in Z\}$$
