1. **Problem statement:** Solve the equation $2 \cos x + 1 = 0$ for $x$ in the interval $0 \leq x < 2\pi$ and find the general solution over all real numbers. 2. **Formula and rules:** We use the fact that $\cos x = y$ can be solved by finding $x = \arccos y$ and using periodicity of cosine: general solution is $x = \pm \arccos y + 2k\pi$, where $k$ is any integer. 3. **Solve the equation:** $$2 \cos x + 1 = 0$$ Subtract 1 from both sides: $$2 \cos x = -1$$ Divide both sides by 2: $$\cancel{2} \cos x = \frac{-1}{\cancel{2}} \implies \cos x = -\frac{1}{2}$$ 4. **Find exact values for $x$ in $0 \leq x < 2\pi$:** Cosine equals $-\frac{1}{2}$ at angles in the second and third quadrants: $$x = \frac{2\pi}{3}, \quad x = \frac{4\pi}{3}$$ 5. **General solution:** Using periodicity of cosine: $$x = \pm \arccos\left(-\frac{1}{2}\right) + 2k\pi = \frac{2\pi}{3} + 2k\pi, \quad \frac{4\pi}{3} + 2k\pi, \quad k \in \mathbb{Z}$$ **Final answers:** - For $0 \leq x < 2\pi$: $x = \frac{2\pi}{3}, \frac{4\pi}{3}$ - General solution: $x = \frac{2\pi}{3} + 2k\pi$ or $x = \frac{4\pi}{3} + 2k\pi$, $k \in \mathbb{Z}$