1. We are asked to solve the equation $\cos^2 x = \frac{3}{4}$ for $0 \leq x \leq 2\pi$. 2. Recall that $\cos^2 x = (\cos x)^2$. So we take the square root of both sides: $$\cos x = \pm \sqrt{\frac{3}{4}} = \pm \frac{\sqrt{3}}{2}$$ 3. The cosine function equals $\frac{\sqrt{3}}{2}$ at angles where $x = \frac{\pi}{6}$ and $x = \frac{11\pi}{6}$ within $[0, 2\pi]$. 4. The cosine function equals $-\frac{\sqrt{3}}{2}$ at angles where $x = \frac{5\pi}{6}$ and $x = \frac{7\pi}{6}$ within $[0, 2\pi]$. 5. Therefore, the solutions are: $$x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$$ This completes the solution for part a.