1. **State the problem:** Solve the equation $5\cos^2 x + 4\cos x = 1$ for $x$. 2. **Rewrite the equation:** Let $y = \cos x$. The equation becomes: $$5y^2 + 4y = 1$$ 3. **Bring all terms to one side:** $$5y^2 + 4y - 1 = 0$$ 4. **Use the quadratic formula:** For $ay^2 + by + c = 0$, the solutions are: $$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Here, $a=5$, $b=4$, $c=-1$. 5. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = 4^2 - 4 \times 5 \times (-1) = 16 + 20 = 36$$ 6. **Find the roots:** $$y = \frac{-4 \pm \sqrt{36}}{2 \times 5} = \frac{-4 \pm 6}{10}$$ 7. **Evaluate each root:** - For $+$ sign: $$y = \frac{-4 + 6}{10} = \frac{2}{10} = 0.2$$ - For $-$ sign: $$y = \frac{-4 - 6}{10} = \frac{-10}{10} = -1$$ 8. **Recall $y = \cos x$:** - $\cos x = 0.2$ - $\cos x = -1$ 9. **Find $x$ values:** - For $\cos x = 0.2$: $$x = \pm \arccos(0.2) + 2k\pi, \quad k \in \mathbb{Z}$$ - For $\cos x = -1$: $$x = \pi + 2k\pi, \quad k \in \mathbb{Z}$$ **Final answer:** $$x = \pm \arccos(0.2) + 2k\pi \quad \text{or} \quad x = \pi + 2k\pi, \quad k \in \mathbb{Z}$$