Answer
$$ x= \frac{\pi}{2}, \frac{3 \pi}{2}, \frac{7 \pi}{6}, \frac{11 \pi}{6}$$
Work Step by Step
We are given $$\sin 2 x+\cos x=0 \text { for } 0 \leq x<2 \pi$$
We factor the left side (and use the double angle formula) as follows:
\begin{align*}
\sin 2 x+\cos x&=0\\
2\sin x\cos x+\cos x&=0\\
\cos x(2\sin x+1)&=0
\end{align*}
Now we set the factors equal to zero
$$ \cos x=0\ \ \ \ \Rightarrow \ \ \ x= \frac{\pi}{2}, \frac{3 \pi}{2}$$
or
$$ 2\sin x+1=0\ \ \ \ \Rightarrow \ \ \ x=\frac{7 \pi}{6}, \frac{11 \pi}{6}$$
Hence the solutions are:
$$ x= \frac{\pi}{2}, \frac{3 \pi}{2}, \frac{7 \pi}{6}, \frac{11 \pi}{6}$$
