Chapter 5 - Section 5.5 - Trigonometric Equations - Exercise Set - Page 703: 42

$$x=0$$

We solve by factoring as follows: $$\cos ^2x+2\cos x -3=0 \quad \Rightarrow \quad (\cos x -1)(\cos x+3)=0 \quad \Rightarrow \quad \cos x -1 =0 \quad \Rightarrow \quad \cos x =1 \quad \Rightarrow \quad x=2n \pi, \quad n \in \mathbb{Z}$$(Please note that the range of $\cos x$ is $[-1, 1]$). So, $x=0$ is the only solution in the interval $[0, 2\pi )$.