1. **State the problem:** Solve the trigonometric equation $$2\cos^2 x - 3\cos x = 0$$ for $x$. 2. **Formula and rules:** This is a quadratic equation in terms of $\cos x$. We can use substitution: let $y = \cos x$. The equation becomes $$2y^2 - 3y = 0$$. 3. **Solve the quadratic equation:** Factor the equation: $$y(2y - 3) = 0$$ This gives two solutions: $$y = 0 \quad \text{or} \quad 2y - 3 = 0 \Rightarrow y = \frac{3}{2}$$ 4. **Check the domain:** Since $y = \cos x$, and $\cos x$ must be in the interval $[-1,1]$, $y = \frac{3}{2}$ is not possible. 5. **Find $x$ for $y=0$:** Solve $$\cos x = 0$$ The general solutions are: $$x = \frac{\pi}{2} + k\pi, \quad k \in \mathbb{Z}$$ 6. **Final answer:** The solutions to the equation are: $$x = \frac{\pi}{2} + k\pi, \quad k \in \mathbb{Z}$$