Precalculus (6th Edition) Blitzer

The solutions of the equation are $\theta =\frac{3\pi }{2}+2n\pi$, where n is any integer.
\begin{align} & 3sin\theta +5=-2\sin \theta \\ & 3sin\theta +2sin\theta =-5 \\ & 5\sin \theta =-5 \\ & \sin \theta =-1 \end{align} $sin\frac{\pi }{2}=1$, and the solutions of $\sin \theta =-1$ in [0,2 $\pi$ ) are: \begin{align} & \theta =\pi +\frac{\pi }{2} \\ & =\frac{2\pi }{2}+\frac{\pi }{2} \\ & =\frac{3\pi }{2} \end{align} \begin{align} & \theta =2\pi -\frac{\pi }{2} \\ & =\frac{4\pi }{2}-\frac{\pi }{2} \\ & =\frac{3\pi }{2} \end{align} The period of the sine function is $2\pi$, and the solutions are given by: $\theta =\frac{3\pi }{2}+2n\pi$, where n is any integer.