## Precalculus (6th Edition) Blitzer

Similarities: Solve $4x+1=3$ by isolating $x$ on one side of the equation: \begin{align} & 4x+1=3 \\ & 4x=2 \\ & x=\frac{1}{2} \end{align} Isolate the $\sin \theta$ from the equation $4\sin \theta +1=3$ \begin{align} & 4\sin \theta =2 \\ & \sin \theta =\frac{1}{2} \\ \end{align} Differences: Isolate the $x$ and solve to get the absolute value for $x$. However, when $\sin \theta$ is isolated, solve for the value of $\theta$ for which $\sin \theta =\frac{1}{2}$; that is, $\sin \theta =\sin \left( \frac{\pi }{6} \right)$ or $\sin \theta =\sin \left( \frac{5\pi }{6} \right)$. So, the values of $\theta$ are $\frac{\pi }{6}$ and $\frac{5\pi }{6}$ between $[0,2\pi )$. Therefore, the statement makes sense.