1. **Problem 3.1:** Find the points of intersection in the first quadrant of the polar curves $r_1 = \sin \theta$ and $r_2 = \sin(2\theta)$.\n\n2. The points of intersection satisfy $r_1 = r_2$, so:\n$$\sin \theta = \sin(2\theta)$$\nRecall the double-angle identity: $\sin(2\theta) = 2 \sin \theta \cos \theta$. Substitute:\n$$\sin \theta = 2 \sin \theta \cos \theta$$\n\n3. Rearrange the equation:\n$$\sin \theta - 2 \sin \theta \cos \theta = 0$$\n$$\sin \theta (1 - 2 \cos \theta) = 0$$\n\n4. Set each factor equal to zero:\n- $\sin \theta = 0$\n- $1 - 2 \cos \theta = 0$\n\n5. Solve each:\n- $\sin \theta = 0$ gives $\theta = 0, \pi, 2\pi, ...$ but in the first quadrant $\theta = 0$ (on boundary).\n- $1 - 2 \cos \theta = 0 \Rightarrow \cos \theta = \frac{1}{2}$\n\n6. For $\cos \theta = \frac{1}{2}$ in the first quadrant, $\theta = \frac{\pi}{3}$.\n\n7. Find $r$ at $\theta = \frac{\pi}{3}$:\n$$r = \sin \left( \frac{\pi}{3} \right) = \frac{\sqrt{3}}{2}$$\n\n8. So the point of intersection in the first quadrant is:\n$$\left( r, \theta \right) = \left( \frac{\sqrt{3}}{2}, \frac{\pi}{3} \right)$$\n\n---\n\n1. **Problem 3.2:** Find the area in the first quadrant outside $r_1$ and inside $r_2$.\n\n2. The area between two polar curves from $\theta = a$ to $\theta = b$ is:\n$$A = \frac{1}{2} \int_a^b \left( r_2^2 - r_1^2 \right) d\theta$$\n\n3. Here, $r_1 = \sin \theta$, $r_2 = \sin(2\theta)$, and the first quadrant corresponds to $\theta \in [0, \frac{\pi}{2}]$.\n\n4. Compute the integral:\n$$A = \frac{1}{2} \int_0^{\frac{\pi}{2}} \left( \sin^2(2\theta) - \sin^2 \theta \right) d\theta$$\n\n5. Use the identity $\sin^2 x = \frac{1 - \cos(2x)}{2}$:\n$$\sin^2(2\theta) = \frac{1 - \cos(4\theta)}{2}, \quad \sin^2 \theta = \frac{1 - \cos(2\theta)}{2}$$\n\n6. Substitute:\n$$A = \frac{1}{2} \int_0^{\frac{\pi}{2}} \left( \frac{1 - \cos(4\theta)}{2} - \frac{1 - \cos(2\theta)}{2} \right) d\theta = \frac{1}{2} \int_0^{\frac{\pi}{2}} \frac{- \cos(4\theta) + \cos(2\theta)}{2} d\theta$$\n\n7. Simplify:\n$$A = \frac{1}{4} \int_0^{\frac{\pi}{2}} \left( \cos(2\theta) - \cos(4\theta) \right) d\theta$$\n\n8. Integrate term-by-term:\n$$\int \cos(2\theta) d\theta = \frac{\sin(2\theta)}{2}, \quad \int \cos(4\theta) d\theta = \frac{\sin(4\theta)}{4}$$\n\n9. Evaluate definite integral:\n$$A = \frac{1}{4} \left[ \frac{\sin(2\theta)}{2} - \frac{\sin(4\theta)}{4} \right]_0^{\frac{\pi}{2}} = \frac{1}{4} \left( \frac{\sin(\pi)}{2} - \frac{\sin(2\pi)}{4} - 0 \right) = 0$$\n\n10. Since the integral evaluates to zero, check the limits of integration. The curves intersect at $\theta = \frac{\pi}{3}$, so split the integral:\n$$A = \frac{1}{2} \left( \int_0^{\frac{\pi}{3}} (r_2^2 - r_1^2) d\theta + \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} (r_2^2 - r_1^2) d\theta \right)$$\n\n11. But the region outside $r_1$ and inside $r_2$ in the first quadrant is from $\theta = 0$ to $\theta = \frac{\pi}{3}$ where $r_2 > r_1$. So the area is:\n$$A = \frac{1}{2} \int_0^{\frac{\pi}{3}} \left( \sin^2(2\theta) - \sin^2 \theta \right) d\theta$$\n\n12. Using the same substitution as before:\n$$A = \frac{1}{4} \int_0^{\frac{\pi}{3}} \left( \cos(2\theta) - \cos(4\theta) \right) d\theta = \frac{1}{4} \left[ \frac{\sin(2\theta)}{2} - \frac{\sin(4\theta)}{4} \right]_0^{\frac{\pi}{3}}$$\n\n13. Evaluate:\n$$= \frac{1}{4} \left( \frac{\sin(\frac{2\pi}{3})}{2} - \frac{\sin(\frac{4\pi}{3})}{4} - 0 \right) = \frac{1}{4} \left( \frac{\sqrt{3}/2}{2} - \frac{-\sqrt{3}/2}{4} \right) = \frac{1}{4} \left( \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{8} \right)$$\n\n14. Simplify:\n$$= \frac{1}{4} \times \frac{3\sqrt{3}}{8} = \frac{3\sqrt{3}}{32}$$\n\n---\n\n1. **Problem 3.3:** Find the slope of the tangent line to $r_2 = \sin(2\theta)$ at $\theta = \frac{\pi}{6}$.\n\n2. The slope $\frac{dy}{dx}$ in polar coordinates is given by:\n$$\frac{dy}{dx} = \frac{\frac{dr}{d\theta} \sin \theta + r \cos \theta}{\frac{dr}{d\theta} \cos \theta - r \sin \theta}$$\n\n3. Compute $r$ and $\frac{dr}{d\theta}$ at $\theta = \frac{\pi}{6}$:\n$$r = \sin(2 \times \frac{\pi}{6}) = \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$$\n$$\frac{dr}{d\theta} = 2 \cos(2\theta) \Rightarrow 2 \cos \frac{\pi}{3} = 2 \times \frac{1}{2} = 1$$\n\n4. Substitute into the slope formula:\n$$\frac{dy}{dx} = \frac{1 \times \sin \frac{\pi}{6} + \frac{\sqrt{3}}{2} \times \cos \frac{\pi}{6}}{1 \times \cos \frac{\pi}{6} - \frac{\sqrt{3}}{2} \times \sin \frac{\pi}{6}}$$\n\n5. Evaluate trigonometric values:\n$$\sin \frac{\pi}{6} = \frac{1}{2}, \quad \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$$\n\n6. Calculate numerator:\n$$1 \times \frac{1}{2} + \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} = \frac{1}{2} + \frac{3}{4} = \frac{5}{4}$$\n\n7. Calculate denominator:\n$$1 \times \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} \times \frac{1}{2} = \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{4}$$\n\n8. Therefore, the slope is:\n$$\frac{dy}{dx} = \frac{\frac{5}{4}}{\frac{\sqrt{3}}{4}} = \frac{5}{4} \times \frac{4}{\sqrt{3}} = \frac{5}{\sqrt{3}} = \frac{5 \sqrt{3}}{3}$$\n\n**Final answers:**\n- 3.1 Intersection point: $\left( \frac{\sqrt{3}}{2}, \frac{\pi}{3} \right)$\n- 3.2 Area: $\frac{3 \sqrt{3}}{32}$\n- 3.3 Slope at $\theta = \frac{\pi}{6}$: $\frac{5 \sqrt{3}}{3}$