1. **State the problem:** Solve the trigonometric equation $$5 \cos^2 x + 4 \cos x = 1$$ for $$0 \leq x \leq 2\pi$$. 2. **Rewrite the equation:** Move all terms to one side to set the equation to zero: $$5 \cos^2 x + 4 \cos x - 1 = 0$$ 3. **Substitute:** Let $$y = \cos x$$. The equation becomes a quadratic: $$5y^2 + 4y - 1 = 0$$ 4. **Solve the quadratic equation:** Use the quadratic formula: $$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$a=5$$, $$b=4$$, and $$c=-1$$. Calculate the discriminant: $$\Delta = 4^2 - 4 \times 5 \times (-1) = 16 + 20 = 36$$ Calculate the roots: $$y = \frac{-4 \pm \sqrt{36}}{2 \times 5} = \frac{-4 \pm 6}{10}$$ So, - $$y_1 = \frac{-4 + 6}{10} = \frac{2}{10} = 0.2$$ - $$y_2 = \frac{-4 - 6}{10} = \frac{-10}{10} = -1$$ 5. **Find $$x$$ values:** Recall $$y = \cos x$$. - For $$\cos x = 0.2$$, $$x = \arccos(0.2)$$. The solutions in $$[0, 2\pi]$$ are: $$x = \arccos(0.2)$$ and $$x = 2\pi - \arccos(0.2)$$. - For $$\cos x = -1$$, $$x = \pi$$. 6. **Final answers:** $$x = \arccos(0.2), \quad x = 2\pi - \arccos(0.2), \quad x = \pi$$ These are the solutions to the equation in the interval $$0 \leq x \leq 2\pi$$.