## Precalculus (10th Edition)

$\theta=\dfrac{\pi}{12}+n\pi$ and $\theta=\dfrac{5\pi}{12}+n\pi$
We are given the equation: $\sin (2\theta)=0.5$ $\sin (2\theta)=\dfrac{1}{2}$ The angles $2\theta$ in $[0,2\pi)$ matching the equation are: $2\theta=\dfrac{\pi}{6}$ and $2\theta=\dfrac{5\pi}{6}$ Given the fact that the function has a period of $2\pi$, the solutions are: $2\theta=\dfrac{\pi}{6}+2n\pi$ and $2\theta=\dfrac{5\pi}{6}+2n\pi$ We determine $\theta$: $\theta=\dfrac{\pi}{12}+n\pi$ and $\theta=\dfrac{5\pi}{12}+n\pi$