## Precalculus (6th Edition) Blitzer

The solution of the given trigonometric equation is $\theta =0,\pi$
We have the equation as $\sin \theta =\tan \theta$. By solving the equation, \begin{align} & \sin \theta =\tan \theta \\ & \sin \theta =\frac{\sin \theta }{\cos \theta } \\ & \sin \theta =\frac{\sin \theta }{\sqrt{1-{{\sin }^{2}}\theta }} \\ & \sin \theta \times \sqrt{1-{{\sin }^{2}}\theta }=\sin \theta \end{align} Then, \begin{align} & \sin \theta \times \sqrt{1-{{\sin }^{2}}\theta }-\sin \theta =0 \\ & \sin \theta \left( \sqrt{1-{{\sin }^{2}}\theta }-1 \right)=0 \end{align} And from the two factors, we get \begin{align} & \sin \theta =0 \\ & \theta =n\pi \end{align} and, \begin{align} & 1-{{\sin }^{2}}\theta =1 \\ & {{\sin }^{2}}\theta =0 \\ & \sin \theta =0 \\ & \theta =n\pi \end{align} Therefore, we get $\theta =n\pi$. The provided range is $\left[ 0,2\pi \right)$. Hence, $\theta =0,\pi$ are the solutions of the provided equation.