1. **Stating the problem:** We want to find the values of $\theta$ for which the tangent to the cardioid given by the polar equation $$r = a(1 + \cos \theta)$$ is parallel to the initial line (the polar axis). 2. **Formula and explanation:** The slope of the tangent line in polar coordinates is given by $$\frac{dy}{dx} = \frac{r' \sin \theta + r \cos \theta}{r' \cos \theta - r \sin \theta}$$ where $$r' = \frac{dr}{d\theta}$$. The tangent is parallel to the initial line (the $x$-axis) when the slope is zero, i.e., when the numerator is zero: $$r' \sin \theta + r \cos \theta = 0$$ 3. **Calculate $r'$:** $$r = a(1 + \cos \theta)$$ $$r' = \frac{dr}{d\theta} = a(-\sin \theta)$$ 4. **Substitute into the numerator condition:** $$a(-\sin \theta) \sin \theta + a(1 + \cos \theta) \cos \theta = 0$$ Simplify: $$-a \sin^2 \theta + a(1 + \cos \theta) \cos \theta = 0$$ Divide both sides by $a$ (assuming $a \neq 0$): $$-\sin^2 \theta + (1 + \cos \theta) \cos \theta = 0$$ 5. **Expand and rearrange:** $$-\sin^2 \theta + \cos \theta + \cos^2 \theta = 0$$ Recall the Pythagorean identity: $$\sin^2 \theta = 1 - \cos^2 \theta$$, substitute: $$-(1 - \cos^2 \theta) + \cos \theta + \cos^2 \theta = 0$$ Simplify: $$-1 + \cos^2 \theta + \cos \theta + \cos^2 \theta = 0$$ $$-1 + 2 \cos^2 \theta + \cos \theta = 0$$ 6. **Rewrite as a quadratic in $x = \cos \theta$:** $$2x^2 + x - 1 = 0$$ 7. **Solve the quadratic:** Using the quadratic formula: $$x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4}$$ So, $$x_1 = \frac{-1 + 3}{4} = \frac{2}{4} = \frac{1}{2}$$ $$x_2 = \frac{-1 - 3}{4} = \frac{-4}{4} = -1$$ 8. **Find corresponding $\theta$ values:** - For $\cos \theta = \frac{1}{2}$, $\theta = \frac{\pi}{3}$ (and also $\theta = \frac{5\pi}{3}$ but not listed). - For $\cos \theta = -1$, $\theta = \pi$. 9. **Check given options:** - $\theta = 0$ (cos 0 = 1) not a root. - $\theta = \pi$ (cos $\pi = -1$) root. - $\theta = \frac{\pi}{3}$ (cos $\frac{\pi}{3} = \frac{1}{2}$) root. - $\theta = \frac{\pi}{2}$ (cos $\frac{\pi}{2} = 0$) not a root. **Final answer:** The tangent is parallel to the initial line when $$\theta = \pi \text{ or } \theta = \frac{\pi}{3}$$.