## Calculus (3rd Edition)

$$x= \frac{\pi}{2}, \frac{3 \pi}{2}, \frac{7 \pi}{6}, \frac{11 \pi}{6}$$
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: $$x= \frac{\pi}{2}, \frac{3 \pi}{2}, \frac{7 \pi}{6}, \frac{11 \pi}{6}$$