1. **State the problem:** Solve the equation $$-4\cos(5x) + 1 = 1$$ for all solutions in radians, where $n$ is any integer. 2. Simplify the equation: \[-4\cos(5x) + 1 = 1\] Subtract 1 from both sides: \[-4\cos(5x) = 0\] Divide both sides by -4: \[\cos(5x) = 0\] 3. Solve $$\cos(5x) = 0$$. The cosine equals zero at angles \[\theta = \frac{\pi}{2} + n\pi, \quad n \in \mathbb{Z}\] Set $$5x = \frac{\pi}{2} + n\pi$$ 4. Solve for $x$: \[x = \frac{\frac{\pi}{2} + n\pi}{5} = \frac{\pi}{10} + n\frac{\pi}{5}\] 5. Write the general solutions: \[x = \frac{\pi}{10} + n\frac{\pi}{5}, \quad n \in \mathbb{Z}\] 6. Check which choices match this general solution. - Choice C: $$ -\frac{\pi}{10} + n \frac{2\pi}{5} $$ (does not match exactly) - Choice D: $$ \frac{\pi}{20} + n \frac{\pi}{5} $$ (does not match exactly) - Choice E: $$ \frac{\pi}{10} + n \frac{2\pi}{5} $$ (similar but with period $$\frac{2\pi}{5}$$ instead of $$\frac{\pi}{5}$$) 7. Note that the principal period of cosine is $2\pi$, so for $$\cos(5x)$$ the period is $$\frac{2\pi}{5}$$, so adding multiples of $$\frac{2\pi}{5}$$ yields all solutions. 8. However, the solution formula from step 4 is $$x = \frac{\pi}{10} + n \frac{\pi}{5}$$, but noting that cosine is zero at $$\frac{\pi}{2} + n\pi$$, the $n$ in the solution covers all integers, so using $$n\frac{2\pi}{5}$$ is correct because $$\frac{\pi}{5}$$ increments will repeat half the period twice. 9. Check choice E, which matches the solutions exactly with proper period: $$x = \frac{\pi}{10} + n \frac{2\pi}{5}$$ 10. Check also choice C: $$x = -\frac{\pi}{10} + n \frac{2\pi}{5}$$ is also valid because cosine is even, and adding period shifts solutions accordingly. 11. Therefore, the solutions are represented by choices C and E. 12. Verify that other choices do not correspond to solution set for $x$. **Final answers:** choices C and E.