Answer
$$x=0, \qquad x=\pi$$
Work Step by Step
We solve by factoring:
$$\cos ^2 \theta -1=0 \quad \Rightarrow \quad \cos ^2 \theta =1 \quad \Rightarrow \quad \cos \theta = \pm 1 \quad \Rightarrow \quad x= 2n\pi, \quad \text{ or } \quad x=\pi + 2n \pi, \quad n \in \mathbb{Z}$$So, $x= 0$ and $x= \pi$ are the only solutions in the interval $[0, 2\pi )$.
