Answer
$$x=\frac{7\pi}{18},\ \frac{11\pi}{18}, \ \frac{19\pi}{18},\ \frac{23\pi}{18},\ \frac{31\pi}{18},\ \frac{35\pi}{18}$$
Work Step by Step
Given $$ \sin 2x \cos x + \cos 2x \sin x=\frac{-1}{2},\ \ 0\leq x<2\pi $$
Then by using low of $\sin(A+B)=\sin A \cos B - \cos A \sin B$ , we get
\begin{align*}
\sin 2x \cos x +\cos 2x \sin x&=\frac{-1}{2}\\
\sin(3x)&=\frac{-1}{2}\\
\end{align*}
First we find all possible solutions for $x$
$$ 3x=\frac{-\pi}{6}+2k\pi \ \Rightarrow x= \frac{-\pi}{18}+ \frac{2k\pi}{3}$$
To find those solutions that lie in the interval $0\leq x<2\pi$ we let k take on values of 0, 1, and 2; we get
$$x=\frac{7\pi}{18},\ \frac{11\pi}{18}, \ \frac{19\pi}{18},\ \frac{23\pi}{18},\ \frac{31\pi}{18},\ \frac{35\pi}{18}$$
