1. **State the problem:** Solve the equation $$\sin x \cos x = 5x^2 \sin x \cos x$$ for $$0 < x < \pi$$. 2. **Rewrite the equation:** The equation is $$\sin x \cos x = 5x^2 \sin x \cos x$$. 3. **Factor the equation:** Move all terms to one side: $$\sin x \cos x - 5x^2 \sin x \cos x = 0$$ Factor out $$\sin x \cos x$$: $$\sin x \cos x (1 - 5x^2) = 0$$ 4. **Set each factor equal to zero:** - $$\sin x = 0$$ - $$\cos x = 0$$ - $$1 - 5x^2 = 0$$ 5. **Solve each equation in the interval $$0 < x < \pi$$:** - $$\sin x = 0$$ at $$x = 0, \pi$$ but $$0$$ and $$\pi$$ are not in the open interval, so no solution here. - $$\cos x = 0$$ at $$x = \frac{\pi}{2}$$ which is in the interval. - $$1 - 5x^2 = 0 \Rightarrow 5x^2 = 1 \Rightarrow x^2 = \frac{1}{5} \Rightarrow x = \pm \frac{1}{\sqrt{5}}$$ Only positive root $$x = \frac{1}{\sqrt{5}}$$ is in $$0 < x < \pi$$. 6. **Final solutions:** $$x = \frac{1}{\sqrt{5}} \approx 0.447$$ radians $$x = \frac{\pi}{2} \approx 1.571$$ radians Both rounded to 3 significant figures. **Answer:** $$x \approx 0.447, 1.57$$ radians.