1. Stating the problem: Solve the equation $$9 \cos(2\theta) + 3 = 5 \cos \theta$$ for $$\theta$$. 2. Recall the double-angle identity for cosine: $$\cos(2\theta) = 2\cos^2\theta - 1$$. 3. Substitute the identity into the equation: $$9(2\cos^2\theta - 1) + 3 = 5 \cos \theta$$ 4. Expand and simplify: $$18\cos^2\theta - 9 + 3 = 5 \cos \theta$$ $$18\cos^2\theta - 6 = 5 \cos \theta$$ 5. Rearrange all terms to one side: $$18\cos^2\theta - 5 \cos \theta - 6 = 0$$ 6. Let $$x = \cos \theta$$, then the quadratic equation is: $$18x^2 - 5x - 6 = 0$$ 7. Solve the quadratic using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \times 18 \times (-6)}}{2 \times 18}$$ $$= \frac{5 \pm \sqrt{25 + 432}}{36} = \frac{5 \pm \sqrt{457}}{36}$$ 8. Calculate the approximate values: $$\sqrt{457} \approx 21.4$$ So, $$x_1 = \frac{5 + 21.4}{36} = \frac{26.4}{36} \approx 0.733$$ $$x_2 = \frac{5 - 21.4}{36} = \frac{-16.4}{36} \approx -0.456$$ 9. Find $$\theta$$ by taking the inverse cosine: $$\theta_1 = \cos^{-1}(0.733) \approx 42.7^\circ$$ $$\theta_2 = \cos^{-1}(-0.456) \approx 117.2^\circ$$ 10. Since cosine is periodic with period $$360^\circ$$, general solutions are: $$\theta = 42.7^\circ + 360^\circ k, \quad \theta = 360^\circ - 42.7^\circ + 360^\circ k$$ $$\theta = 117.2^\circ + 360^\circ k, \quad \theta = 360^\circ - 117.2^\circ + 360^\circ k$$ for any integer $$k$$. Final answer: $$\theta \approx 42.7^\circ, 317.3^\circ, 117.2^\circ, 242.8^\circ + 360^\circ k$$