1. **State the problem:** Solve the inequality $$\sin x \cos^3 x > \cos x \sin^3 x$$ for $$x \in ]0, 2\pi[$$. 2. **Rewrite the inequality:** $$\sin x \cos^3 x > \cos x \sin^3 x$$ can be rearranged as $$\sin x \cos^3 x - \cos x \sin^3 x > 0$$ 3. **Factor the expression:** Factor out $$\sin x \cos x$$: $$\sin x \cos x (\cos^2 x - \sin^2 x) > 0$$ 4. **Use trigonometric identities:** Recall that $$\cos^2 x - \sin^2 x = \cos 2x$$, so the inequality becomes $$\sin x \cos x \cos 2x > 0$$ 5. **Analyze the sign of each factor:** We want $$\sin x > 0$$ $$\cos x > 0$$ $$\cos 2x > 0$$ because the product of three terms is positive only if an even number of them are negative, but since all three multiplied is positive, and the factors are continuous, we consider the intervals where all three are positive or exactly two are negative (which would make product positive). But it's easier to analyze sign intervals. 6. **Determine intervals for $$\sin x > 0$$:** $$\sin x > 0$$ on $$]0, \pi[$$ 7. **Determine intervals for $$\cos x > 0$$:** $$\cos x > 0$$ on $$]-\frac{\pi}{2}, \frac{\pi}{2}[$$ and $$]\frac{3\pi}{2}, \frac{5\pi}{2}[$$ Within $$]0, 2\pi[$, this is $$]0, \frac{\pi}{2}[$$ and $$]\frac{3\pi}{2}, 2\pi[$$ 8. **Determine intervals for $$\cos 2x > 0$$:** Since $$\cos 2x$$ has period $$\pi$$, it is positive on intervals $$\left(-\frac{\pi}{2} + k\pi, \frac{\pi}{2} + k\pi\right)$$ for integer $$k$$. Within $$]0, 2\pi[$, these intervals are: - $$]0, \frac{\pi}{2}[$ (for $$k=0$$) - $$]\pi, \frac{3\pi}{2}[$ (for $$k=1$$) 9. **Find intersection of intervals where product is positive:** We want $$\sin x \cos x \cos 2x > 0$$. Check intervals where all three factors are positive: - $$\sin x > 0$$ on $$]0, \pi[$ - $$\cos x > 0$$ on $$]0, \frac{\pi}{2}[$ and $$]\frac{3\pi}{2}, 2\pi[$ - $$\cos 2x > 0$$ on $$]0, \frac{\pi}{2}[$ and $$]\pi, \frac{3\pi}{2}[$ The intersection of all three positive intervals is: - $$]0, \frac{\pi}{2}[$ (since all three positive here) Check intervals where exactly two factors are negative (product positive): - $$\sin x > 0$$ on $$]0, \pi[$, so $$\sin x$$ negative on $$]\pi, 2\pi[$ - $$\cos x > 0$$ on $$]0, \frac{\pi}{2}[$ and $$]\frac{3\pi}{2}, 2\pi[$, so negative on $$]\frac{\pi}{2}, \frac{3\pi}{2}[$ - $$\cos 2x > 0$$ on $$]0, \frac{\pi}{2}[$ and $$]\pi, \frac{3\pi}{2}[$, so negative on $$]\frac{\pi}{2}, \pi[$ and $$]\frac{3\pi}{2}, 2\pi[$ Check intervals where product is positive due to two negatives: - On $$]\frac{\pi}{2}, \pi[$: - $$\sin x > 0$$ (positive) - $$\cos x < 0$$ (negative) - $$\cos 2x < 0$$ (negative) Two negatives, product positive. - On $$]\frac{3\pi}{2}, 2\pi[$: - $$\sin x < 0$$ (negative) - $$\cos x > 0$$ (positive) - $$\cos 2x < 0$$ (negative) Two negatives, product positive. 10. **Combine all intervals where product is positive:** $$]0, \frac{\pi}{2}[ \cup ]\frac{\pi}{2}, \pi[ \cup ]\frac{3\pi}{2}, 2\pi[$ Simplify: $$]0, \pi[ \cup ]\frac{3\pi}{2}, 2\pi[$ 11. **Exclude points where any factor is zero:** - $$\sin x = 0$$ at $$0, \pi, 2\pi$$ - $$\cos x = 0$$ at $$\frac{\pi}{2}, \frac{3\pi}{2}$$ - $$\cos 2x = 0$$ at $$\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$$ So intervals are open intervals excluding these points. 12. **Match with given options:** Option (d) is: $$]0, \frac{\pi}{2}[ \cup ]\frac{3\pi}{2}, \pi[ \cup ]\frac{3\pi}{2}, \frac{7\pi}{4}[$ But $$]\frac{3\pi}{2}, \pi[$ is empty since $$\pi < \frac{3\pi}{2}$$, so likely a typo. Option (a) is: $$]0, \frac{\pi}{4}[ \cup ]\frac{\pi}{2}, \frac{3\pi}{4}[ \cup ]\pi, \frac{5\pi}{4}[ \cup ]\frac{3\pi}{2}, \frac{7\pi}{4}[$ This matches the intervals where $$\cos 2x = 0$$ at $$\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$$ split the domain. From the analysis, the solution set is: $$]0, \frac{\pi}{2}[ \cup ]\frac{3\pi}{2}, 2\pi[$ which corresponds best to option (d) if corrected. **Final answer:** The solution set is $$]0, \frac{\pi}{2}[ \cup ]\frac{3\pi}{2}, 2\pi[$.