## University Calculus: Early Transcendentals (3rd Edition)

$$\theta=\Big[\frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3}\Big]$$
$$\sin^2\theta=\frac{3}{4}$$ where $0\le\theta\le2\pi$ We take the square root: $$\sin\theta=\pm\frac{\sqrt3}{2}$$ - For $\sin\theta=\frac{\sqrt3}{2}$: Since $\theta\in[0,2\pi]$, which is the whole unit circle, there are 2 points where $\sin\theta=\frac{\sqrt3}{2}$, which are $\theta=\frac{\pi}{3}$ and $\theta=\frac{2\pi}{3}$. - For $\sin\theta=-\frac{\sqrt3}{2}$: Since $\theta\in[0,2\pi]$, which is the whole unit circle, there are 2 points where $\sin\theta=-\frac{\sqrt3}{2}$, which are $\theta=\frac{4\pi}{3}$ and $\theta=\frac{5\pi}{3}$. Therefore, overall, the answer to this exercise is $$\theta=\Big[\frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3}\Big]$$