Answer
The solutions of the equation are $\theta =\frac{3\pi }{2}+2n\pi $, where n is any integer.
Work Step by Step
$\begin{align}
& 3sin\theta +5=-2\sin \theta \\
& 3sin\theta +2sin\theta =-5 \\
& 5\sin \theta =-5 \\
& \sin \theta =-1
\end{align}$
$sin\frac{\pi }{2}=1$, 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.
