Answer
The solution set is $$\{360^\circ n,60^\circ+360^\circ n,180^\circ+360^\circ n,300^\circ+360^\circ n,n\in Z\}$$
Work Step by Step
$$\sin\theta-\sin2\theta=0$$
- Recall the identity: $\sin2\theta=2\sin\theta\cos\theta$ and apply it to the given equation:
$$\sin\theta-2\sin\theta\cos\theta=0$$
$$\sin\theta(1-2\cos\theta)=0$$
$$\sin\theta=0\hspace{1cm}\text{or}\hspace{1cm}\cos\theta=\frac{1}{2}$$
1) First, we solve the equation over the interval $[0^\circ,360^\circ)$
- For $\sin\theta=0$, over the interval $[0^\circ, 360^\circ)$, there are two values of $\theta$ where $\sin\theta=0$, which are $\theta=\{0^\circ,180^\circ\}$.
- For $\cos\theta=\frac{1}{2}$, over the interval $[0^\circ, 360^\circ)$, there are also two values of $\theta$ where $\cos\theta=\frac{1}{2}$, which are $\theta=\{60^\circ,300^\circ\}$.
Therefore, $$\theta=\{0^\circ,60^\circ,180^\circ,300^\circ\}$$
2) Solve the equation for all solutions
Both sine function and cosine function have period $360^\circ$, so we would add $360^\circ$ to all solutions found in part 1) for $\theta$.
$$\theta=\{0^\circ+360^\circ n,60^\circ+360^\circ n,180^\circ+360^\circ n,300^\circ+360^\circ n,n\in Z\}$$
$$\theta=\{360^\circ n,60^\circ+360^\circ n,180^\circ+360^\circ n,300^\circ+360^\circ n,n\in Z\}$$
This is also the solution set.
