1. **State the problem:** Solve the trigonometric equation $$2\cos^2(x) - 3\cos(x) + 1 = 0$$ for $$x$$ in the interval $$[0, 2\pi]$$. 2. **Identify the formula and substitution:** Let $$y = \cos(x)$$. The equation becomes a quadratic in $$y$$: $$2y^2 - 3y + 1 = 0$$. 3. **Solve the quadratic equation:** Use the quadratic formula: $$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$a=2$$, $$b=-3$$, and $$c=1$$. Calculate the discriminant: $$\Delta = (-3)^2 - 4 \times 2 \times 1 = 9 - 8 = 1$$. So, $$y = \frac{3 \pm \sqrt{1}}{4} = \frac{3 \pm 1}{4}$$. 4. **Find the roots:** - $$y_1 = \frac{3 + 1}{4} = 1$$ - $$y_2 = \frac{3 - 1}{4} = \frac{2}{4} = 0.5$$ 5. **Back-substitute to find $$x$$:** - For $$\cos(x) = 1$$, $$x = 0$$ (within $$[0, 2\pi]$$). - For $$\cos(x) = 0.5$$, $$x = \pm \frac{\pi}{3} + 2k\pi$$. Within $$[0, 2\pi]$$, the solutions are: - $$x = \frac{\pi}{3}$$ - $$x = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$$ 6. **Final answer:** $$x = 0, \frac{\pi}{3}, \frac{5\pi}{3}$$ These are the solutions to the equation in the interval $$[0, 2\pi]$$.