1. **State the problem:** Solve the differential equation $$y''(x)\cos(2x) + y'(x) = 0.$$ 2. **Rewrite the equation:** Let $y' = p$, then $y'' = p'$. The equation becomes $$p'(x)\cos(2x) + p(x) = 0.$$ 3. **Isolate $p'$:** $$p'(x) = -\frac{p(x)}{\cos(2x)}.$$ 4. **Separate variables:** $$\frac{dp}{p} = -\frac{dx}{\cos(2x)}.$$ 5. **Integrate both sides:** $$\int \frac{1}{p} dp = -\int \frac{1}{\cos(2x)} dx.$$ 6. **Left integral:** $$\int \frac{1}{p} dp = \ln|p| + C_1.$$ 7. **Right integral:** Recall $$\frac{1}{\cos(2x)} = \sec(2x).$$ So $$\int \sec(2x) dx = \frac{1}{2} \ln|\sec(2x) + \tan(2x)| + C_2.$$ 8. **Combine integrals:** $$\ln|p| = -\frac{1}{2} \ln|\sec(2x) + \tan(2x)| + C,$$ where $C = C_2 - C_1.$ 9. **Exponentiate both sides:** $$|p| = e^C \cdot \left|\sec(2x) + \tan(2x)\right|^{-\frac{1}{2}}.$$ Let $A = e^C.$ 10. **Recall $p = y'$:** $$y' = A \left(\sec(2x) + \tan(2x)\right)^{-\frac{1}{2}}.$$ 11. **Integrate to find $y$:** $$y = A \int \left(\sec(2x) + \tan(2x)\right)^{-\frac{1}{2}} dx + B,$$ where $B$ is a constant of integration. **Final answer:** $$y = A \int \left(\sec(2x) + \tan(2x)\right)^{-\frac{1}{2}} dx + B.$$ This integral may require special functions or numerical methods for explicit evaluation.