1. **State the problem:** Solve the equation $$2(\sin x)^2 - 5 \cos x + 1 = 0$$ for $$x$$ in the interval $$[0, 2\pi]$$, expressing answers as multiples of $$\pi$$. 2. **Use the Pythagorean identity:** Recall that $$\sin^2 x = 1 - \cos^2 x$$. 3. **Rewrite the equation:** Substitute $$\sin^2 x$$: $$2(1 - \cos^2 x) - 5 \cos x + 1 = 0$$ 4. **Simplify:** $$2 - 2 \cos^2 x - 5 \cos x + 1 = 0$$ $$-2 \cos^2 x - 5 \cos x + 3 = 0$$ 5. **Multiply both sides by $$-1$$ to simplify:** $$2 \cos^2 x + 5 \cos x - 3 = 0$$ 6. **Let $$y = \cos x$$, then solve the quadratic:** $$2 y^2 + 5 y - 3 = 0$$ 7. **Use the quadratic formula:** $$y = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 2 \cdot (-3)}}{2 \cdot 2} = \frac{-5 \pm \sqrt{25 + 24}}{4} = \frac{-5 \pm \sqrt{49}}{4}$$ 8. **Calculate roots:** $$y_1 = \frac{-5 + 7}{4} = \frac{2}{4} = \frac{1}{2}$$ $$y_2 = \frac{-5 - 7}{4} = \frac{-12}{4} = -3$$ 9. **Check domain:** Since $$\cos x$$ must be between $$-1$$ and $$1$$, discard $$y_2 = -3$$. 10. **Solve $$\cos x = \frac{1}{2}$$ in $$[0, 2\pi]$$:** $$x = \frac{\pi}{3}, \frac{5\pi}{3}$$ 11. **Express answers as multiples of $$\pi$$:** $$x = \frac{1}{3} \pi, \frac{5}{3} \pi$$ **Final answer:** $$\boxed{\frac{1}{3}, \frac{5}{3}}$$