Answer
$$\theta=\Big[\frac{\pi}{6},\frac{\pi}{2},\frac{5\pi}{6},\frac{3\pi}{2}\Big]$$
Work Step by Step
$$\sin2\theta-\cos\theta=0$$
where $0\le\theta\le2\pi$
Apply the Double Angle Formula here for $\sin2\theta$, which states $\sin2\theta=2\sin\theta\cos\theta$
That means, $$2\sin\theta\cos\theta-\cos\theta=0$$
$$\cos\theta(2\sin\theta-1)=0$$
$$\cos\theta=0\hspace{1cm}\text{or}\hspace{1cm}\sin\theta=\frac{1}{2}$$
- For $\cos\theta=0$:
As $\theta\in[0,2\pi]$, there are 2 points where $\cos\theta=0$, which are $\theta=\frac{\pi}{2}$ and $\theta=\frac{3\pi}{2}$.
- For $\sin\theta=\frac{1}{2}$:
As $\theta\in[0,2\pi]$, there are also 2 points where $\sin\theta=\frac{1}{2}$, which are $\theta=\frac{\pi}{6}$ and $\theta=\frac{5\pi}{6}$.
Therefore, overall, the answer to this exercise is
$$\theta=\Big[\frac{\pi}{6},\frac{\pi}{2},\frac{5\pi}{6},\frac{3\pi}{2}\Big]$$
