Answer
The solutions are $\theta =\frac{3\pi }{4},\frac{7\pi }{4}$.
Work Step by Step
Solve equation (1):
$\begin{align}
& \sin \theta \cos \theta =-\frac{1}{2} \\
& \frac{\sin 2\theta }{2}=-\frac{1}{2} \\
& \sin 2\theta =-1
\end{align}$
As the given period is $2\pi $, therefore the value on which $\sin \theta =-1$ is $\frac{3\pi }{2}$.
Now, $2\theta =\frac{3\pi }{2}$.
As the given period is $2\pi $, all the solutions to $\sin 2\theta =-1$ are:
$\begin{align}
& 2\theta =\frac{3\pi }{2}+2n\pi \\
& \theta =\frac{3\pi }{4}+n\pi
\end{align}$
Therefore, the solution in $\left[ 0,\left. 2\pi \right) \right.$ can be obtained by letting $ n=0$ and $ n=1$, such that:
$\theta =\frac{3\pi }{4},\frac{7\pi }{4}$
